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Sequences and Series

3.3 Convergence Tests

53 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Select the series below that diverge by the divergence test.

(A)  ∑n=1∞1n\displaystyle\sum_{n=1}^\infty \frac{1}{n} (B)  ∑n=1∞n2n+1\displaystyle\sum_{n=1}^\infty \frac{n^2}{n+1} (C) ∑n=1∞sin⁡n\displaystyle\sum_{n=1}^\infty \sin n (D)  ∑n=1∞sin⁡(πn)\displaystyle\sum_{n=1}^\infty \sin (\pi n)

Hint

That is, which series have terms whose limit is not zero?

Answer

(B), (C)

Full solution

  1. lim⁡n→∞1n=0\displaystyle\lim_{n \to \infty}\frac{1}{n}=0, so the divergence test is inconclusive. It's true that this series diverges, but we can't show it using the divergence test.

  2. lim⁡n→∞n2n+1=∞\displaystyle\lim_{n \to \infty}\frac{n^2}{n+1}=\infty, which is not zero, so the divergence test tells us this series diverges.

  3. We'll show below that lim⁡n→∞sin⁡n\displaystyle\lim_{n \to \infty}\sin n does not exist at all. In particular, it is not zero. Therefore, the divergence test tells us this series diverges.

    Now we'll show that lim⁡n→∞sin⁡n\displaystyle\lim_{n \to \infty}\sin n does not exist. Suppose that it does exist and takes the value SS. We will now see that this assumption leads to a contradiction. Add together the two trig identities (see Appendix A.8 in the CLP-2 text)

    sin⁡(n+1)=sin⁡(n)cos⁡(1)+cos⁡(n)sin⁡(1)sin⁡(n−1)=sin⁡(n)cos⁡(1)−cos⁡(n)sin⁡(1)\begin{align*} \sin(n+1) &= \sin(n)\cos(1)+\cos(n)\sin(1) \\ \sin(n-1) &= \sin(n)\cos(1)-\cos(n)\sin(1) \end{align*}

    This gives

    sin⁡(n+1)+sin⁡(n−1)=2sin⁡(n)cos⁡(1)\begin{equation*} \sin(n+1) +\sin (n-1) = 2\sin(n)\cos(1) \end{equation*}

    Taking the limit n→∞n\rightarrow\infty gives 2S=2Scos⁡(1)2S=2S\cos(1). Since cos⁡(1)≠1\cos(1)\ne 1, this forces S=0S=0. Now the first trig identity above gives

    cos⁡(n)=sin⁡(n+1)−sin⁡(n)cos⁡(1)sin⁡(1)\begin{equation*} \cos(n) = \frac{\sin(n+1)-\sin(n)\cos(1)}{\sin(1)} \end{equation*}

    Taking the limit as n→∞n\rightarrow\infty of that gives

    lim⁡n→∞cos⁡(n)=S−Scos⁡(1)sin⁡(1)=0\begin{equation*} \lim_{n\to\infty}\cos(n) = \frac{S-S\cos(1)}{\sin(1)}=0 \end{equation*}

    But that provides the contradiction. Because sin⁡2(n)+cos⁡2(n)=1\sin^2(n)+\cos^2(n)=1, we can't have both sin⁡(n)\sin(n) and cos⁡(n)\cos(n) converging to zero. So lim⁡n→∞sin⁡n\displaystyle\lim_{n \to \infty}\sin n does not exist.

  4. For all whole numbers nn, sin⁡(πn)=0\sin(\pi n)=0, so lim⁡n→∞sin⁡(πn)=0\displaystyle\lim_{n \to \infty}\sin(\pi n)=0 and the divergence test is inconclusive.

Q2Stage 1

Select the series below whose terms satisfy the conditions to apply the integral test.

(A)  ∑n=1∞1n\displaystyle\sum_{n=1}^\infty \frac{1}{n} (B)  ∑n=1∞n2n+1\displaystyle\sum_{n=1}^\infty \frac{n^2}{n+1} (C) ∑n=1∞sin⁡n\displaystyle\sum_{n=1}^\infty \sin n (D)  ∑n=1∞sin⁡n+1n2\displaystyle\sum_{n=1}^\infty \frac{\sin n+1}{n^2}

Hint

That is, if f(x)f(x) is a function with f(n)=anf(n)=a_n for all whole numbers nn, is f(x)f(x) nonnegative and decreasing?

Answer

(A)

Full solution

Let f(x)f(x) be a function with f(n)=anf(n)=a_n for all whole numbers nn. In order to apply the integral test (Theorem 3.3.5 in the CLP-2 text) we need f(x)f(x) to be positive and decreasing for all sufficiently large values of nn.

  1. f(x)=1xf(x) = \frac{1}{x}, which is positive and decreasing for all x≥1x \ge 1, so the integral test does apply here.

  2. f(x)=x2x+1f(x)=\frac{x^2}{x+1}, which is not decreasing–in fact, it goes to infinity. So, the integral test does not apply here. (The divergence test tells us the series diverges, though.)

  3. f(x)=sin⁡xf(x) = \sin x, which is neither consistently positive nor consistently decreasing, so the integral test does not apply. (The divergence test tells us the series diverges, though.)

  4. f(x)=sin⁡x+1x2f(x)=\frac{\sin x+1}{x^2} is positive for all whole numbers nn. To determine whether it is decreasing, we consider its derivative.

    f′(x)=x2(cos⁡x)−(sin⁡x+1)(2x)x4=xcos⁡x−2sin⁡x−2x3\begin{align*} f'(x)&=\frac{x^2(\cos x)-(\sin x+1)(2x)}{x^4}=\frac{x\cos x - 2\sin x -2}{x^3} \end{align*}

    This is sometimes positive, and sometimes negative. (For example, if x=100πx=100\pi, f′(x)=100π−0−2(100π)3>0f'(x) = \frac{100\pi-0-2}{(100\pi)^3}>0, but if x=101πx=101\pi then f′(x)=101π(−1)−0−2(101π)3<0f'(x)=\frac{101\pi(-1)-0-2}{(101\pi)^3}<0.) Then f(x)f(x) is not a decreasing function, so the integral test does not apply.

Q3Stage 1

Suppose there is some threshold after which a person is considered old, and before which they are young.

Let Olaf be an old person, and let Yuan be a young person.

  1. Suppose I am older than Olaf. Am I old?

  2. Suppose I am younger than Olaf. Am I old?

  3. Suppose I am older than Yuan. Am I young?

  4. Suppose I am younger than Yuan. Am I young?

Hint

This isn't a trick. It's meant to give you intuition to the direct comparison test.

Answer

(a) I am old (b) not enough information to tell
(c) not enough information to tell (d) I am young

Full solution

(a) If Olaf is old, and I am even older, then I am old as well.
(b) If Olaf is old, and I am not as old, then perhaps I am old as well (just slightly less so), or perhaps I am young. There is not enough information to tell.
(c) If Yuan is young, and I am older, then perhaps I am much older and I am old, or perhaps I am only a little older, and I am young. There is not enough information to tell.
(d) If Yuan is young, and I am even younger, then I must also be young.

Another way to think about this is with a timeline of birthdates. People born before the threshold are old, and people born after it are young.

Figure from prob_s3.3, line 1

Figure from prob_s3.3, line 1

If I'm born before (older than) Olaf, I'm born before the threshold, so I'm old.
If I'm born after (younger than) Yuan I'm born after the threshold, so I'm young.

Figure from prob_s3.3, line 1

Figure from prob_s3.3, line 1

If I'm born after Olaf or before Yuan, I don't know which side of the threshold I'm on. I could be old or I could be young.

Figure from prob_s3.3, line 1

Figure from prob_s3.3, line 1

Q4Stage 1

Below are graphs of two sequences with positive terms. Assume the sequences continue as shown. Fill in the table with conclusions that can be made from the direct comparison test, if any.

Figure from prob_s3.3, line 1

Figure from prob_s3.3, line 1

if ∑12an\sum\limits^{\vphantom{\frac12}} a_n convergesif ∑an\sum a_n diverges
[10pt] and if {an}\{a_n\} is the red seriesthen ∑12bn\sum\limits^{\vphantom{\frac12}} b_n \rule1.5cm1ptthen ∑bn\sum b_n \rule1.5cm1pt
[10pt] and if {an}\{a_n\} is the blue seriesthen ∑12bn\sum\limits^{\vphantom{\frac12}} b_n \rule1.5cm1ptthen ∑bn\sum b_n \rule1.5cm1pt
[10pt]
Hint

The comparison test is Theorem 3.3.8 in the CLP-2 text. However, rather than trying to memorize which way the inequalities go in all cases, you can use the same reasoning as Question 3.

Answer
if ∑12an\sum\limits^{\vphantom{\frac12}} a_n convergesif ∑an\sum a_n diverges
[10pt] and if {an}\{a_n\} is the red seriesthen ∑12bn\sum\limits^{\vphantom{\frac12}} b_n CONVERGESinconclusive
[10pt] and if {an}\{a_n\} is the blue seriesinconclusivethen ∑12bn\sum\limits^{\vphantom{\frac12}} b_n DIVERGES
[10pt]
Full solution

The comparison test is Theorem 3.3.8 in the CLP-2 text. However, rather than trying to memorize which way the inequalities go in all cases, we use the same reasoning as Question 3.

If a sequence has positive terms, it either converges, or it diverges to infinity, with the partial sums increasing and increasing without bound. If one sequence diverges, and the other sequence is larger, then the other sequence diverges–just like being older than an old person makes you old.

If ∑an\sum a_n converges, and {an}\{a_n\} is the red (larger) series, then ∑bn\sum b_n converges: it's smaller than a sequence that doesn't add up to infinity, so it too does not add up to infinity.

If ∑an\sum a_n diverges, and {an}\{a_n\} is the blue (smaller) series, then ∑bn\sum b_n diverges: it's larger than a sequence that adds up to infinity, so it too adds up to infinity.

In the other cases, we can't say anything. If {an}\{a_n\} is the red (larger) series, and ∑an\sum a_n diverges, then perhaps {bn}\{b_n\} behaves similarly to {an}\{a_n\} and ∑bn\sum b_n diverges, or perhaps {bn}\{b_n\} is much, much smaller than {an}\{a_n\} and ∑bn\sum b_n converges.

Similarly, if {an}\{a_n\} is the blue (smaller) series, and ∑an\sum a_n converges, then perhaps {bn}\{b_n\} behaves similarly to {an}\{a_n\} and ∑bn\sum b_n converges, or perhaps {bn}\{b_n\} is much, much bigger than {an}\{a_n\} and ∑bn\sum b_n diverges.

if ∑12an\sum\limits^{\vphantom{\frac12}} a_n convergesif ∑an\sum a_n diverges
[10pt] and if {an}\{a_n\} is the red seriesthen ∑12bn\sum\limits^{\vphantom{\frac12}} b_n CONVERGESinconclusive
[10pt] and if {an}\{a_n\} is the blue seriesinconclusivethen ∑12bn\sum\limits^{\vphantom{\frac12}} b_n DIVERGES
[10pt]
Q5Stage 1

For each pair of series below, decide whether the second series is a valid comparison series to determine the convergence of the first series, using the direct comparison test and/or the limit comparison test.

  1. ∑n=10∞1n−1,\displaystyle\sum_{n=10}^{\infty} \frac{1}{n-1}, compared to the divergent series ∑n=10∞1n.\displaystyle\sum_{n=10}^{\infty} \frac{1}{n}.

  2. ∑n=1∞sin⁡nn2+1,\displaystyle\sum_{n=1}^{\infty} \frac{\sin n}{n^2+1}, compared to the convergent series ∑n=1∞1n2.\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}.

  3. ∑n=5∞n3+5n+1n6−2,\displaystyle\sum_{n=5}^{\infty} \frac{n^3+5n+1}{n^6-2}, compared to the convergent series ∑n=5∞1n3.\displaystyle\sum_{n=5}^{\infty} \frac{1}{n^3}.

  4. ∑n=5∞1n,\displaystyle\sum_{n=5}^{\infty} \frac{1}{\sqrt{n}}, compared to the divergent series ∑n=5∞1n4.\displaystyle\sum_{n=5}^{\infty} \frac{1}{\sqrt[4]n}.

Hint

Think about Question 4 to remind yourself which way the inequalities have to go for direct comparison.

Note that all the comparison series have positive terms, so we don't need to worry about that part of the limit comparison test.

Answer

(a) both direct comparison and limit comparison (b) direct comparison
(c) limit comparison (d) neither

Full solution

(a) Since ∑1n\sum \frac{1}{n} is divergent, we can only use it to prove series with larger terms are divergent. This is the case here, since 1n−1>1n\frac{1}{n-1}>\frac{1}{n}. So, the direct comparison test is valid.

For the limit comparison test, we calculate:

lim⁡n→∞anbn=lim⁡n→∞1n−11n=lim⁡n→∞11−1n=1\begin{align*} \lim_{n \to \infty}\frac{a_n}{b_n}&=\lim_{n \to \infty}\frac{\frac{1}{n-1}}{\frac{1}{n}} =\lim_{n \to \infty}\frac{1}{1-\frac{1}{n}} =1 \end{align*}

Since the limit exists and is not zero, the limit comparison test is also valid.

(b) Since the series ∑1n2\sum \frac{1}{n^2} converges, we can only use the direct comparison test to show the convergence of a series if its terms have smaller absolute values. Indeed,

∣sin⁡nn2+1∣=∣sin⁡n∣n2+1<1n2\left| \frac{\sin n}{n^2+1}\right|=\frac{|\sin n|}{n^2+1}<\frac{1}{n^2}

so the series are set for a direct comparison.

To check whether a limit comparison will work, we compute:

lim⁡n→∞anbn=lim⁡n→∞sin⁡nn2+11n2=lim⁡n→∞n2n2+1sin⁡n=lim⁡n→∞(1)sin⁡n\begin{align*} \lim_{n \to \infty}\frac{a_n}{b_n}&=\lim_{n \to \infty} \frac{\frac{\sin n}{n^2+1}}{\frac{1}{n^2}} = \lim_{n \to \infty}\frac{n^2}{n^2+1}\sin n = \lim_{n \to \infty}(1)\sin n \end{align*}

The limit does not exist, so the limit comparison test is not a valid test to compare these two series.

(c) Since the series ∑1n3\sum \frac{1}{n^3} converges, we can only use the direct comparison test to conclude something about a series with smaller terms. However,

n3+5n+1n6−2>n3n6=1n3.\frac{n^3+5n+1}{n^6-2}>\frac{n^3}{n^6}=\frac{1}{n^3}.

Therefore the direct comparison test does not apply to this pair of series.

For the limit comparison test, we calculate:

lim⁡n→∞anbn=lim⁡n→∞n3+5n+1n6−21n3=lim⁡n→∞n3+5n+1n3−2n3(1n31n3)=lim⁡n→∞1+5n2+1n31−2n6=1\begin{align*} \lim_{n \to \infty}\frac{a_n}{b_n}&=\lim_{n \to \infty}\frac{\frac{n^3+5n+1}{n^6-2}}{ \frac{1}{n^3}} = \lim_{n \to \infty}\frac{n^3+5n+1}{n^3-\frac{2}{n^3}}\left(\frac{\frac{1}{n^3}}{\frac{1}{n^3}}\right)\\ &=\lim_{n \to \infty}\frac{1+\frac{5}{n^2}+\frac{1}{n^3}}{1-\frac{2}{n^6}}=1 \end{align*}

Since the limit is a nonzero real number, we can use the limit comparison test to compare this pair of series.

(d) Since the series ∑1n4\sum \frac{1}{\sqrt[4]{n}} diverges, we can only use the direct comparison test to show that a series with larger terms diverges. However,

1n<1n4\frac{1}{\sqrt{n}}<\frac{1}{\sqrt[4]{n}}

so the direct comparison test isn't valid with this pair of series.

For the limit comparison test, we calculate:

L=lim⁡n→∞anbn=lim⁡n→∞1n1n4=lim⁡n→∞1n4=0\begin{align*} L=\lim_{n \to \infty}\frac{a_n}{b_n}&=\lim_{n \to \infty}\frac{\frac{1}{\sqrt{n}}}{\frac{1}{\sqrt[4]{n}}}=\lim_{n \to \infty}\frac{1}{\sqrt[4]{n}}=0 \end{align*}

Since the series ∑1n4\sum \frac{1}{\sqrt[4]{n}} diverges, it is part (b) of the limit comparison test, Theorem 3.3.11 in the CLP-2 text, that is appropriate here. As the limit L=0L=0, the limit comparison test doesn't apply.

Q6Stage 1

Suppose ana_n is a sequence with lim⁡n→∞an=12\displaystyle\lim_{n \to \infty}a_n = \frac{1}{2}. Does ∑n=7∞an\displaystyle\sum_{n=7}^\infty a_n converge or diverge, or is it not possible to determine this from the information given? Why?

Hint

The divergence test is Theorem 3.3.1 in the CLP-2 text.

Answer

It diverges by the divergence test, because lim⁡n→∞an≠0\displaystyle\lim_{n \to \infty}a_n \neq 0.

Full solution

It diverges by the divergence test, because lim⁡n→∞an≠0\displaystyle\lim_{n \to \infty}a_n \neq 0.

Q7Stage 1

What flaw renders the following reasoning invalid?

Q: Determine whether ∑n=1∞sin⁡nn\displaystyle\sum_{n=1}^\infty \dfrac{\sin n}{n} converges or diverges.
A: First, we will evaluate lim⁡n→∞sin⁡nn\displaystyle\lim_{n \to \infty} \dfrac{\sin n}{n}.

  • Note −1n≤sin⁡nn≤1n\dfrac{-1}{n} \leq \dfrac{\sin n}{n} \leq \dfrac{1}{n} for n≥1n \ge 1.

  • Note also that lim⁡n→∞−1n=lim⁡n→∞1n=0\displaystyle\lim_{n \to \infty}\frac{-1}{n}=\displaystyle\lim_{n \to \infty}\frac{1}{n}=0.

  • Therefore, by the Squeeze Theorem, lim⁡n→∞sin⁡nn=0\displaystyle\lim_{n \to \infty} \dfrac{\sin n}{n}=0 as well.

So, by the divergence test, ∑n=1∞sin⁡nn\displaystyle\sum_{n=1}^\infty \dfrac{\sin n}{n} converges.

Hint

The limit is calculated correctly.

Answer

We cannot use the divergence test to show that a series converges. It is inconclusive in this case.

Full solution

The divergence test (Theorem 3.3.1 in the CLP-2 text) is inconclusive when lim⁡n→∞an=0\displaystyle\lim_{n \to \infty}a_n=0. We cannot use the divergence test to show that a series converges.

Q8Stage 1

What flaw renders the following reasoning invalid?

Q: Determine whether ∑n=1∞(sin⁡(πn)+2)\displaystyle\sum_{n=1}^\infty \left(\sin(\pi n)+2\right) converges or diverges.
A: We use the integral test. Let f(x)=sin⁡(πx)+2f(x)=\sin(\pi x)+2. Note f(x)f(x) is always positive, since sin⁡(x)+2≥−1+2=1\sin(x)+2 \geq -1+2 =1. Also, f(x)f(x) is continuous.

∫1∞[sin⁡(πx)+2]dx=lim⁡b→∞∫1b[sin⁡(πx)+2]dx=lim⁡b→∞[−1πcos⁡(πx)+2x∣1b]=lim⁡b→∞[−1πcos⁡(πb)+2b+1π(−1)−2]=∞\begin{align*} \int_1^\infty [\sin(\pi x)+2] dx &= \lim_{b \to \infty}\int_1^b [\sin(\pi x)+2 ]dx\\ &=\lim_{b \to \infty} \left[\left.-\frac{1}{\pi}\cos(\pi x)+2x \right|_1^b\right]\\ &=\lim_{b \to \infty}\left[ -\frac{1}{\pi}\cos(\pi b)+2b +\frac{1}{\pi}(-1)-2\right]\\ &=\infty \end{align*}

By the integral test, since the integral diverges, also ∑n=1∞(sin⁡(πn)+2)\displaystyle\sum_{n=1}^\infty\left( \sin(\pi n)+2\right) diverges.

Hint

It is true that f(x)f(x) is positive. What else has to be true of f(x)f(x) for the integral test to apply?

Answer

The integral test does not apply because f(x)f(x) is not decreasing.

Full solution

The integral test does not apply because f(x)f(x) is not decreasing.

Q9Stage 1

What flaw renders the following reasoning invalid?

Q: Determine whether the series ∑n=1∞2n+1n2en+2n\displaystyle\sum_{n=1}^\infty \dfrac{2^{n+1}n^2}{e^n+2n} converges or diverges.
A: We want to compare this series to the series ∑n=1∞2n+1en\displaystyle\sum_{n=1}^\infty \dfrac{2^{n+1}}{e^n}. Note both this series and the series in the question have positive terms.

First, we find that 2n+1n2en+2n>2n+1en\dfrac{2^{n+1}n^2}{e^n+2n} > \dfrac{2^{n+1}}{e^n} when nn is sufficiently large. The justification for this claim is as follows:

  • We note that en(n2−1)>n2−1>2ne^n(n^2-1)>n^2-1>2n for nn sufficiently large.

  • Therefore, en⋅n2>en+2ne^n \cdot n^2 > e^n+2n

  • Therefore, 2n+1⋅en⋅n2>2n+1(en+2n)2^{n+1}\cdot e^n \cdot n^2 > 2^{n+1}(e^n+2n)

  • Since en+2ne^n+2n and ene^n are both expressions that work out to be positive for the values of nn under consideration, we can divide both sides of the inequality by these terms without having to flip the inequality. So, 2n+1n2en+2n>2n+1en\dfrac{2^{n+1}n^2}{e^n+2n}>\dfrac{2^{n+1}}{e^n}.

Now, we claim ∑n=1∞2n+1en\displaystyle\sum_{n=1}^\infty \dfrac{2^{n+1}}{e^n} converges.
Note ∑n=1∞2n+1en=2∑n=1∞2nen=2∑n=1∞(2e)n\displaystyle\sum_{n=1}^\infty \dfrac{2^{n+1}}{e^n}= 2\displaystyle\sum_{n=1}^\infty \dfrac{2^{n}}{e^n}= 2\displaystyle\sum_{n=1}^\infty \left(\dfrac{2}{e}\right)^n. This is a geometric series with r=2er=\frac{2}{e}. Since 2/e<12/e <1, the series converges.

Now, by the Direct Comparison Test, we conclude that ∑n=1∞2n+1n2en+2n\displaystyle\sum_{n=1}^\infty \dfrac{2^{n+1}n^2}{e^n+2n} converges.

Hint

Refer to Question 4.

Answer

The inequality goes the wrong way, so the direct comparison test (with this comparison series) is inconclusive.

Full solution

The inequality goes the wrong way, so the direct comparison test (with this comparison series) is inconclusive.

Q10Stage 1

Which of the series below are alternating?

(A) ∑n=1∞sin⁡n\displaystyle\sum_{n=1}^\infty \sin n (B) ∑n=1∞cos⁡(πn)n3\displaystyle\sum_{n=1}^\infty \frac{\cos(\pi n)}{n^3} (C) ∑n=1∞7(−n)2n\displaystyle\sum_{n=1}^\infty \frac{7}{(-n)^{2n}} (D) ∑n=1∞(−2)n3n+1\displaystyle\sum_{n=1}^\infty \frac{(-2)^n}{3^{n+1}}

Hint

The definition of an alternating series is given in the start of Section 3.3.4 in the CLP-2 text.

Answer

(B), (D)

Full solution

Although the terms of (A) are sometimes negative and sometimes positive, they are not strictly alternating in a positive-negative-positive-negative pattern. For instance, sin⁡1\sin 1 and sin⁡2\sin 2 are both postive. So, (A) is not an alternating series.

When nn is a whole number, cos⁡(πn)=(−1)n\cos(\pi n) = (-1)^n, so (B) is alternating.

Since the exponent of (−n)(-n) in (C) is even, the terms are always positive. Therefore (C) is not alternating.

(D) is an alternating series.

Q11Stage 1

Give an example of a convergent series for which the ratio test is inconclusive.

Hint

For the ratio test to be inconclusive, lim⁡n→∞∣an+1an∣\lim\limits_{n \to \infty}\left|\dfrac{a_{n+1}}{a_n}\right| should be 1 or nonexistent.

Answer

One possible answer: ∑n=1∞1n2\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}.

Full solution

One possible answer: ∑n=1∞1n2\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}. This series converges (it's a pp–series with p=2>1p=2>1), but if we take the ratio of consecutive terms:

lim⁡n→∞an+1an=lim⁡n→∞n2(n+1)2=1\lim_{n \to \infty}\frac{a_{n+1}}{a_n} = \lim_{n \to \infty}\frac{n^2}{(n+1)^2}=1

The limit of the ratio is 1, so the ratio test is inconclusive.

Q12Stage 1

Imagine you're taking an exam, and you momentarily forget exactly how the inequality in the ratio test works. You remember there's a ratio, but you don't remember which term goes on top; you remember there's something about the limit being greater than or less than one, but you don't remember which way implies convergence.

Explain why

lim⁡n→∞∣an+1an∣>1\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|>1

or, equivalently,

lim⁡n→∞∣anan+1∣<1\lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right|<1

should mean that the sum ∑n=1∞an\sum\limits_{n=1}^\infty a_n diverges (rather than converging).

Hint

By the divergence test, for a series ∑an\sum a_n to converge, we need lim⁡n→∞an=0\lim\limits_{n \to \infty} a_n=0. That is, the magnitude (absolute value) of the terms needs to be getting smaller.

Answer

By the divergence test, for a series ∑an\sum a_n to converge, we need lim⁡n→∞an=0\lim\limits_{n \to \infty} a_n=0. That is, the magnitude (absolute value) of the terms needs to be getting smaller. If lim⁡n→∞∣anan+1∣<1\displaystyle\lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right|<1 or (equivalently) lim⁡n→∞∣an+1an∣>1\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|>1, then ∣an+1∣>∣an∣|a_{n+1}|>|a_n| for sufficiently large nn, so the terms are actually growing in magnitude. That means the series diverges, by the divergence test.

Full solution

By the divergence test, for a series ∑an\sum a_n to converge, we need lim⁡n→∞an=0\lim\limits_{n \to \infty} a_n=0. That is, the magnitude (absolute value) of the terms needs to be getting smaller. If lim⁡n→∞∣anan+1∣<1\displaystyle\lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right|<1 or (equivalently) lim⁡n→∞∣an+1an∣>1\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|>1, then ∣an+1∣>∣an∣|a_{n+1}|>|a_n| for sufficiently large nn, so the terms are actually growing in magnitude. That means the series diverges, by the divergence test.

Q13Stage 1

Give an example of a series ∑n=a∞an\displaystyle\sum_{n=a}^\infty a_n, with a function f(x)f(x) such that f(n)=anf(n)=a_n for all whole numbers nn, such that:

  • ∫a∞f(x) dx\displaystyle\int_a^\infty f(x)\,\dee{x} diverges, while

  • ∑n=a∞an\displaystyle\sum_{n=a}^\infty a_n converges.

Hint

If f(x)f(x) is positive and decreasing, then the integral test tells you that the integral and the series either both increase or both decrease. So, in order to find an example with the properties required in the question, you need f(x)f(x) to not be both positive and decreasing.

Answer

One possible answer: f(x)=sin⁡(πx)f(x) = \sin(\pi x), an=0a_n=0 for every nn.

By the integral test, any answer will use a function f(x)f(x) that is not both positive and decreasing.

Full solution

The terms of the series only see a small portion of the domain of the integral. We can try to think of a function f(x)f(x) that behaves “nicely" when xx is a whole number (that is, it produces a sequence whose sum converges), but is more unruly when xx is not a whole number.

For example, suppose f(x)=sin⁡(πx)f(x)=\sin(\pi x). Then f(x)=0f(x)=0 for every integer xx, but this is not representative of the function as a whole. Indeed, our corresponding series has terms {an}={0,0,0,…}\{a_n\}=\{0,0,0,\ldots\}.

  • ∫1∞sin⁡(πx) dx=lim⁡R→∞[−1πcos⁡(πx)]1R=lim⁡R→∞[−cos⁡(πR)]−1π\displaystyle\int_{1}^\infty \sin(\pi x)\,\dee{x} = \lim_{R \to \infty} \left[-\frac{1}{\pi}\cos (\pi x)\right]_1^R = \lim_{R \to \infty}\Big[-\cos (\pi R)\Big]-\frac{1}{\pi}
    Since the limit does not exist, the integral diverges.

  • ∑n=1∞sin⁡(πn)=∑n=1∞0=0\displaystyle\sum_{n=1}^{\infty}\sin(\pi n) = \sum_{n=1}^\infty 0 = 0. The series converges.

Q14Stage 1Past exam · 2016Q5

Suppose that you want to use the Limit Comparison Test on the series ∑n=0∞an\displaystyle \sum_{n=0}^{\infty} a_n where an=2n+n3n+1\displaystyle a_n = \frac{2^n+n}{3^n+1}. Write down a sequence {bn}\{b_n\} such that lim⁡n→∞anbn\displaystyle \lim\limits_{n\to\infty} \frac{a_n}{b_n} exists and is nonzero. (You don't have to carry out the Limit Comparison Test)

Hint

Review Theorem 3.3.11 and Example 3.3.12 in the CLP-2 text.

Answer

One possible answer: bn=2n3nb_n=\displaystyle \frac{2^n}{3^n}

Full solution

When nn is very large, the term 2n2^n dominates the numerator, and the term 3n3^n dominates the denominator. So when nn is very large an≈2n3na_n\approx \frac{2^n}{3^n}. Therefore we should take bn=2n3nb_n=\displaystyle \frac{2^n}{3^n}. Note that, with this choice of bnb_n,

lim⁡n→∞anbn=lim⁡n→∞2n+n3n+1 3n2n=lim⁡n→∞1+n/2n1+1/3n=1\begin{align*} \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{2^n+n}{3^n+1}\ \frac{3^n}{2^n} = \lim_{n\to\infty} \frac{1+ n/2^{n}}{1+1/3^{n}} =1 \end{align*}

as desired.

Q15Stage 1Past exam · 2014D, 2016A

Decide whether each of the following statements is true or false. If false, provide a counterexample. If true provide a brief justification.

  1. If lim⁡n→∞an=0\displaystyle\lim_{n\rightarrow\infty}a_n=0, then ∑n=1∞an\sum\limits_{n=1}^{\infty} a_n converges.

  2. If lim⁡n→∞an=0\displaystyle\lim_{n\rightarrow\infty}a_n=0, then ∑n=1∞(−1)n(an\sum\limits_{n=1}^{\infty} (-1)^{n\mathstrut} a_n converges.

  3. If 0≤an≤bn0\le a_n \le b_n and ∑n=1∞bn\sum\limits_{n=1}^{\infty} b_n diverges, then ∑n=1∞an\sum\limits_{n=1}^{\infty} a_n diverges.

Hint

Don't jump to conclusions about properties of the ana_n's.

Answer

(a) In general false. The harmonic series ∑n=1∞1n\sum\limits_{n=1}^\infty\frac{1}{n} provides a counterexample.

(b) In general false. If an=(−1)n1na_n =(-1)^n\frac{1}{n}, then ∑n=1∞(−1)n(an\sum\limits_{n=1}^{\infty} (-1)^{n\mathstrut} a_n is again the harmonic series ∑n=1∞1n\sum_{n=1}\limits^\infty\frac{1}{n}, which diverges.

(c) In general false. Take, for example, an=0a_n=0 and bn=1b_n=1.

Full solution

(a) In general false. The harmonic series ∑n=1∞1n\sum\limits_{n=1}^\infty\frac{1}{n} diverges by the pp–test with p=1p=1.

(b) Be careful. You were not told that the ana_n's are positive. So this is false in general. If an=(−1)n1na_n =(-1)^n\frac{1}{n}, then ∑n=1∞(−1)n(an\sum\limits_{n=1}^{\infty} (-1)^{n\mathstrut} a_n is again the harmonic series ∑n=1∞1n\sum\limits_{n=1}^\infty\frac{1}{n}, which diverges.

(c) In general false. Take, for example, an=0a_n=0 and bn=1b_n=1.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q16Stage 2Past exam · 2016Q5

Does the series ∑n=2∞n23n2+n\displaystyle \sum_{n=2}^\infty \frac{n^2}{3n^2+\sqrt n} converge?

Hint

Always try the divergence test first (in your head).

Answer

No. It diverges.

Full solution

First, we'll check the divergence test. It doesn't always work, but if it does, it's likely the easiest path.

lim⁡n→∞n23n2+n(1n21n2)=lim⁡n→∞13+1nn=13≠0\begin{align*} \lim_{n\to\infty} \frac{n^2}{3n^2+\sqrt n} \left(\frac{\frac{1}{n^2}}{\frac{1}{n^2}}\right) = \lim_{n\to\infty} \frac{1}{3+\frac{1}{n\sqrt n}} = \frac{1}{3} \neq 0 \end{align*}

Since the limit of the terms being added is not zero, the series diverges by the divergence test.

Q17Stage 2Past exam · 2014D

Determine, with explanation, whether the series ∑n=1∞5k4k+3k\displaystyle \sum_{n=1}^\infty \frac{5^k}{4^k+3^k} converges or diverges.

Hint

Which test should you always try first (in your head)?

Answer

It diverges.

Full solution

This precise question was asked on a 2014 final exam. Note that the nthn^{\rm th} term in the series is an=5k4k+3ka_n=\frac{5^k}{4^k+3^k} and does not depend on nn! There are two possibilities. Either this was intentional (and the instructor was being particularly nasty) or it was a typo and the intention was to have an=5n4n+3na_n=\frac{5^n}{4^n+3^n}. In both cases, the limit

lim⁡n→∞an=lim⁡n→∞5k4k+3k=5k4k+3k≠0lim⁡n→∞an=lim⁡n→∞5n4n+3n=lim⁡n→∞(5/4)n1+(3/4)n=+∞≠0\begin{align*} \lim_{n\to\infty} a_n &=\lim_{n\to\infty} \frac{5^k}{4^k+3^k} = \frac{5^k}{4^k+3^k} \neq 0 \\ \lim_{n\to\infty} a_n &=\lim_{n\to\infty} \frac{5^n}{4^n+3^n} =\lim_{n\to\infty} \frac{(5/4)^n}{1+(3/4)^n} = +\infty \neq 0 \\ \end{align*}

is nonzero, so the series diverges by the divergence test.

Q18Stage 2Past exam · 2016Q5

Determine whether the series ∑n=0∞1n+12\displaystyle\sum_{n=0}^\infty\frac{1}{n+\frac{1}{2}} is convergent or divergent. If it is convergent, find its value.

Hint

Review the integral test, which is Theorem 3.3.5 in the CLP-2 text.

Answer

The series diverges.

Full solution

We usually check the divergence test first, to look for low-hanging fruit. The limit of the terms being added is zero:

lim⁡n→∞1n+12=0\lim_{n \to \infty}\frac{1}{n+\frac{1}{2}}=0

so the divergence test is inconclusive. That is, we need to look harder.

Next, we might consider a comparison test–these can also provide us (if we're lucky) with an easy path. The terms we're adding look somewhat like 1n\frac{1}{n}, but our terms are smaller than these terms, which form the terms of the divergent harmonic series. So, a direct comparison seems unlikely. Now we search for more exotic tests.

Let f(x)=1x+12\displaystyle f(x) =\frac{1}{x+\frac{1}{2}}. Note f(x)f(x) is positive and decreases as xx increases. So, by the integral test, which is Theorem 3.3.5 in the CLP-2 text, the given series converges if and only if the integral ∫0∞1x+12 dx\int_0^\infty\frac{1}{x+\frac{1}{2}} \,\dee{x} converges. Since

∫0∞1x+12 dx=lim⁡R→∞∫0R1x+12 dx=lim⁡R→∞[log⁡(x+12)]x=0x=R=lim⁡R→∞[log⁡(R+12)−log⁡12]\begin{align*} \int_0^\infty \frac{1}{x+\frac{1}{2}} \,\dee{x} = \lim_{R\rightarrow\infty} \int_0^R\frac{1}{x+\frac{1}{2}} \,\dee{x} &=\lim_{R\rightarrow\infty} \bigg[ \log\Big(x+\frac{1}{2}\Big)\bigg]^{x=R}_{x=0} \\ &=\lim_{R\to\infty} \bigg[ \log\Big(R+\frac{1}{2}\Big) - \log\frac12 \bigg] \end{align*}

diverges, the series diverges.

Q19Stage 2

Does the following series converge or diverge? ∑k=1∞1kk+1\displaystyle\sum_{k=1}^\infty\frac{1}{\sqrt{k}\sqrt{k+1}}

Hint

A comparison might be helpful–try some algebraic manipulation to find a likely series to compare it to.

Answer

It diverges.

Full solution

The terms of the series tend to 0, so we can't use the divergence test.

To generate a guess about its convergence, we do the following:

∑1kk+1=∑1k2+k≈∑1k2=∑1k\sum \frac{1}{\sqrt{k}\sqrt{k+1}}=\sum\frac{1}{\sqrt{k^2+k}} \approx \sum\frac{1}{\sqrt{k^2}}=\sum\frac{1}{k}

We guess that our series behaves like the harmonic series, and the harmonic series diverges (which can be demonstrated by pp-test or integral test). So, we guess that our series diverges. However, in order to directly compare our series to the harmonic series and show our series diverges, our terms would have to be bigger than the terms in the harmonic series, and this is not the case. So, we use limit comparison.

1k1k2+k=k2+kk=k2+kk2=k2+kk2=1+1k, solim⁡k→∞1k1k2+k=lim⁡k→∞1+1k=1\begin{align*} \frac{\frac{1}{k}}{\frac{1}{\sqrt{k^2+k}}}&=\frac{\sqrt{k^2+k}}{k}=\frac{\sqrt{k^2+k}}{\sqrt{k^2}}= \sqrt{\frac{k^2+k}{k^2}}=\sqrt{1+\frac{1}{k}},\text{ so}\\ \lim_{k \rightarrow \infty}\frac{\frac{1}{k}}{\frac{1}{\sqrt{k^2+k}}}&=\lim_{k \rightarrow \infty}\sqrt{1+\frac{1}{k}}=1 \end{align*}

Since 1 is a real number greater than 0, by the Limit Comparison Test, ∑1kk+1\sum \frac{1}{\sqrt{k}\sqrt{k+1}} diverges, like ∑1k\sum \frac{1}{k}.

Q20Stage 2

Evaluate the following series, or show that it diverges: ∑k=30∞3(1.001)k\displaystyle\sum_{k=30}^\infty 3(1.001)^k.

Hint

This is a geometric series.

Answer

This is a geometric series with r=1.001r=1.001. Since ∣r∣>1|r|>1, it is divergent.

Full solution

This is a geometric series with r=1.001r=1.001. Since ∣r∣>1|r|>1, it is divergent.

Q21Stage 2

Evaluate the following series, or show that it diverges: ∑n=3∞(−15)n\displaystyle\sum_{n=3}^\infty \left(\frac{-1}{5}\right)^n.

Hint

Notice that the series is geometric, but it doesn't start at n=0n=0.

Answer

The series converges to −1150-\dfrac{1}{150}.

Full solution

This is a geometric series with r=−15r=\frac{-1}{5}. Since ∣r∣<1|r|<1, it is convergent.

We want to use the formula ∑n=0∞rn=11−r\sum_{n=0}^\infty r^n=\frac{1}{1-r}, but our series does not start at 0, so we re-write it:

∑n=3∞(−15)n=∑n=0∞(−15)n−∑n=02(−15)n=11−(−1/5)−(1−15+125)=16/5−1+15−125=56+−25+5−125=−1150\begin{align*} \sum_{n=3}^\infty \left(\frac{-1}{5}\right)^n&= \sum_{n=0}^\infty \left(\frac{-1}{5}\right)^n- \sum_{n=0}^2 \left(\frac{-1}{5}\right)^n =\frac{1}{1-(-1/5)} - \left(1-\frac{1}{5}+\frac{1}{25} \right)\\ &=\frac{1}{6/5}-1+\frac{1}{5}-\frac{1}{25} = \frac{5}{6}+\frac{-25+5-1}{25}=-\frac{1}{150} \end{align*}
Q22Stage 2

Does the following series converge or diverge? ∑n=7∞sin⁡(πn)\displaystyle\sum_{n=7}^\infty \sin(\pi n)

Hint

Note nn only takes integer values: what's sin⁡(πn)\sin(\pi n) when nn is an integer?

Answer

The series converges.

Full solution

For any integer nn, sin⁡(πn)=0\sin(\pi n)=0, so ∑sin⁡(πn)=∑0=0\sum \sin(\pi n)=\sum 0 = 0. So, this series converges.

Q23Stage 2

Does the following series converge or diverge? ∑n=7∞cos⁡(πn)\displaystyle\sum_{n=7}^\infty\cos(\pi n)

Hint

Note nn only takes integer values: what's cos⁡(πn)\cos(\pi n) when nn is an integer?

Answer

It diverges.

Full solution

For any integer nn, cos⁡(πn)=±1\cos(\pi n)=\pm 1, so lim⁡n→∞cos⁡(πn)≠0\lim\limits_{n \rightarrow \infty}\cos(\pi n) \neq 0.
By the divergence test, this series diverges.

Q24Stage 2

Does the following series converge or diverge? ∑k=1∞ekk!\displaystyle\sum_{k=1}^\infty \frac{e^k}{k!}.

Hint

What's the test that you should always think of when you see a factorial?

Answer

The series converges.

Full solution

Factorials grow super fast. Like, wow, really fast. Even faster than exponentials. So the terms are going to zero, and the divergence test won't help us. Let's use ratio–it's a good go-to test with factorials.

ak+1ak=ek+1(k+1)!ekk!=ek+1ek⋅k!(k+1)!=e⋅k(k−1)⋯(1)(k+1)(k)(k−1)⋯(1)=e⋅1k+1=ek+1\begin{align*}\frac{a_{k+1}}{a_k}&=\frac{\frac{e^{k+1}}{(k+1)!}}{\frac{e^k}{k!}}=\frac{e^{k+1}}{e^k}\cdot\frac{k!}{(k+1)!} = e\cdot\frac{k(k-1)\cdots(1)}{(k+1)(k)(k-1)\cdots(1)}=e\cdot\frac{1}{k+1}=\frac{e}{k+1}\end{align*}

Since ee is a constant,

lim⁡k→∞ak+1ak=lim⁡k→∞ek+1=0\begin{align*}\lim_{k \rightarrow \infty}\frac{a_{k+1}}{a_k}&=\lim_{k \rightarrow \infty}\frac{e}{k+1}=0\end{align*}

Since 0<10<1, by the ratio test, the series converges.

Q25Stage 2

Evaluate the following series, or show that it diverges: ∑k=0∞2k3k+2\displaystyle\sum_{k=0}^\infty\frac{2^k}{3^{k+2}}.

Hint

This is a geometric series, but you'll need to do a little algebra to figure out rr.

Answer

The series converges to 13\dfrac{1}{3}.

Full solution

This is close to being in the form of a geometric series. First, we should have our powers be kk, not k+2k+2, but we notice 3k+2=3k32=9⋅323^{k+2}=3^k3^2=9\cdot3^2, so:

∑k=0∞2k3k+2=∑k=0∞2k9⋅3k=19∑k=0∞2k3k=19∑k=0∞(23)k\begin{align*}\sum_{k=0}^\infty\frac{2^k}{3^{k+2}}&=\sum_{k=0}^\infty\frac{2^k}{9\cdot3^k} =\frac{1}{9}\sum_{k=0}^\infty\frac{2^k}{3^k}=\frac{1}{9}\sum_{k=0}^\infty\left(\frac{2}{3}\right)^k\end{align*}

Now it looks like a geometric series with r=23r=\frac{2}{3}

=19(11−(2/3))=13\begin{align*}&=\frac{1}{9}\left(\frac{1}{1-(2/3)}\right)=\frac{1}{3}\end{align*}

In conclusion: this (geometric) series is convergent, and its sum is 13\dfrac{1}{3}.

Q26Stage 2

Does the following series converge or diverge? ∑n=1∞n!n!(2n)!\displaystyle\sum_{n=1}^\infty\frac{n!n!}{(2n)!}.

Hint

Which test fits most often with factorials?

Answer

The series converges.

Full solution

Usually with factorials, we want to use the divergence test or the ratio test. Since the terms are indeed tending towards zero, we are left with the ratio test.

an+1an=(n+1)!(n+1)!(2n+2)!n!n!(2n)!=(n+1)!(n+1)!n!n!⋅(2n)!(2n+2)!=(n+1)(n)(n−1)⋯(1)n(n−1)⋯(1)⋅(n+1)(n)(n−1)⋯(1)n(n−1)⋯(1)⋅(2n)(2n−1)(2n−2)⋯(1)(2n+2)(2n+1)(2n)(2n−1)(2n−2)⋯(1)=(n+1)(n+1)⋅1(2n+2)(2n+1), solim⁡n→∞an+1an=lim⁡n→∞(n+1)(n+1)(2n+2)(2n+1)=lim⁡n→∞(n+1)(n+1)2(n+1)(2n+1)=lim⁡n→∞n+14n+2=14\begin{align*} \frac{a_{n+1}}{a_n}&=\frac{\frac{(n+1)!(n+1)!}{(2n+2)!}}{\frac{n!n!}{(2n)!}}=\frac{(n+1)!(n+1)!}{n!n!}\cdot\frac{(2n)!}{(2n+2)!}\\ &=\tfrac{\textcolor{red}{(n+1)}{(n)}{(n-1)}\cdots{(1)}}{{n}{(n-1)}\cdots{(1)}} \cdot\tfrac{\textcolor{red}{(n+1)}{(n)}{(n-1)}\cdots{(1)}}{{n}{(n-1)}\cdots{(1)}} \cdot \tfrac{{(2n)}{(2n-1)}{(2n-2)}\cdots{(1)}}{\textcolor{red}{(2n+2)(2n+1)}{(2n)}{(2n-1)}{(2n-2)}\cdots{(1)}} \\ &=(n+1)(n+1) \cdot \frac{1}{(2n+2)(2n+1)}\text{, so}\\ \lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n}&=\lim_{n \rightarrow \infty}\frac{(n+1)(n+1)}{(2n+2)(2n+1)} =\lim_{n \rightarrow \infty}\frac{(n+1)(n+1)}{2(n+1)(2n+1)} =\lim_{n \rightarrow \infty}\frac{n+1}{4n+2}=\frac{1}{4} \end{align*}

Since the limit is a number less than 1, the series converges by the ratio test.

Q27Stage 2

Does the following series converge or diverge? ∑n=1∞n2+12n4+n\displaystyle\sum_{n=1}^\infty\frac{n^2+1}{2n^4+n}.

Hint

Try finding a nice comparison.

Answer

It converges.

Full solution

We want to make an estimation, when nn gets big:

n2+12n4+n≈n22n4=12n2\frac{n^2+1}{2n^4+n} \approx \frac{n^2}{2n^4}=\frac{1}{2n^2}

Since ∑12n2\sum \frac{1}{2n^2} is a convergent series (by pp-test, or integral test), we guess that our series is convergent as well. If we wanted to use comparison test, we should have to show n2+12n4+n<12n2\frac{n^2+1}{2n^4+n}<\frac{1}{2n^2}, which seems unpleasant, so let's use limit comparison.

lim⁡n→∞n2+12n4+n12n2=lim⁡n→∞(n2+1)2n22n4+n=lim⁡n→∞2n4+2n22n4+n(1/n41/n4)=lim⁡n→∞2+2n22+1n3=1\lim_{n \rightarrow \infty}\frac{\frac{n^2+1}{2n^4+n}}{\frac{1}{2n^2}}=\lim_{n \rightarrow \infty}\frac{(n^2+1)2n^2}{2n^4+n} =\lim_{n \rightarrow \infty}\frac{2n^4+2n^2}{2n^4+n}\left(\frac{1/n^4}{1/n^4}\right)=\lim_{n \rightarrow \infty}\frac{2+\frac{2}{n^2}}{2+\frac{1}{n^3}}=1

Since the limit is a positive finite number, by the Limit Comparison Test, ∑n1+12n4+n\sum \frac{n^1+1}{2n^4+n} does the same thing ∑12n2\sum\frac{1}{2n^2} does: it converges.

Q28Stage 2Past exam · 2016Q5

Show that the series ∑n=3∞5n(log⁡n)3/2\displaystyle\sum_{n=3}^\infty \frac{5}{n(\log n)^{3/2}} converges.

Hint

With the substitution u=log⁡xu=\log x, the function 1x(log⁡x)3/2\dfrac{1}{x(\log x)^{3/2}} is easily integrable.

Answer

Let f(x)=5x(log⁡x)3/2\displaystyle f(x) = \frac{5}{x(\log x)^{3/2}}. Then f(x)f(x) is positive and decreases as xx increases. So the sum ∑3∞f(n)\displaystyle\sum_3^{\infty} f(n) and the integral ∫3∞f(x) dx\displaystyle\int_3^\infty f(x) \,\dee{x} either both converge or both diverge, by the integral test, which is Theorem 3.3.5 in the CLP-2 text. For the integral, we use the substitution u=log⁡xu = \log x, du=dxx\dee{u} = \frac{\dee{x}}{x} to get

∫3∞5 dxx(log⁡x)3/2=∫log⁡3∞5 duu3/2\begin{align*} \int_3^\infty \frac{5 \,\dee{x}}{x(\log x)^{3/2}} = \int_{\log 3}^\infty \frac{5 \,\dee{u}}{u^{3/2}} \end{align*}

which converges by the pp–test (which is Example 1.12.8 in the CLP-2 text) with p=32>1p=\frac{3}{2} > 1.

Full solution

First, we rule out some of the easier tests. The limit of the terms being added is zero, so the divergence test is inconclusive. The terms being added are smaller than the terms of the (divergent) harmonic series, ∑1n\sum \frac{1}{n}, so we can't directly compare these two series, and there isn't another obvious series to compare ours to. However, the terms being added seem like a function we could integrate.

Let f(x)=5x(log⁡x)3/2\displaystyle f(x) = \frac{5}{x(\log x)^{3/2}}. Then f(x)f(x) is positive and decreases as xx increases. So the sum ∑3∞f(n)\displaystyle\sum_3^{\infty} f(n) and the integral ∫3∞f(x) dx\displaystyle\int_3^\infty f(x) \,\dee{x} either both converge or both diverge, by the integral test, which is Theorem 3.3.5 in the CLP-2 text. For the integral, we use the substitution u=log⁡xu = \log x, du=dxx\dee{u} = \frac{\dee{x}}{x} to get

∫3∞5 dxx(log⁡x)3/2=∫log⁡3∞5 duu3/2\begin{align*} \int_3^\infty \frac{5 \,\dee{x}}{x(\log x)^{3/2}} = \int_{\log 3}^\infty \frac{5 \,\dee{u}}{u^{3/2}} \end{align*}

which converges by the pp–test (which is Example 1.12.8 in the CLP-2 text) with p=32>1p=\frac{3}{2} > 1.

Q29Stage 2Past exam · 2014A

Find the values of pp for which the series ∑n=2∞1n(log⁡n)p\displaystyle{\sum_{n=2}^\infty \frac{1}{n(\log n)^p}} converges.

Hint

Combine the integral test with the results about pp-series, Example 3.3.6 in the CLP-2 text.

Answer

p>1p>1

Full solution

Let f(x)=1x(log⁡x)pf(x)=\frac{1}{x(\log x)^p}. Then f(x)f(x) is positive for n≥3n \ge 3, and f(x)f(x) decreases as xx increases. So, we can use the integral test, Theorem 3.3.5 in the CLP-2 text.

∫2∞ ⁣ ⁣1x(log⁡x)pdx=lim⁡R→∞∫2R ⁣ ⁣1(log⁡x)pdxx=lim⁡R→∞∫log⁡2log⁡R ⁣ ⁣1upduwith u=log⁡x, du=dxx\begin{align*} \int_2^\infty\!\! \frac{1}{x(\log x)^p} \dee{x} &=\lim_{R\rightarrow\infty}\int_2^R\!\! \frac{1}{(\log x)^p} \frac{\dee{x}}{x} =\lim_{R\rightarrow\infty}\int_{\log 2}^{\log R}\!\! \frac{1}{u^p} \dee{u} \quad\text{with }u=\log x\text{, }\dee{u}=\frac{\dee{x}}{x} \\ \end{align*}

Using the results about pp-series, Example 3.3.6 in the CLP-2 text, we know this integral converges if and only if p>1p>1, so the same is true for the series by the integral test.

Q30Stage 2Past exam · 2016A

Does ∑n=1∞e−nn{\displaystyle\sum_{n=1}^\infty\frac{e^{-\sqrt{n}}}{\sqrt{n}}} converge or diverge?

Hint

Try the substitution u=xu=\sqrt{x}.

Answer

It converges.

Full solution

As usual, let's see whether the “easy" tests work. The terms we're adding converge to zero:

lim⁡n→∞e−nn=lim⁡n→∞1nen=0\lim_{n \to \infty}\frac{e^{-\sqrt{n}}}{\sqrt{n}}=\lim_{n \to \infty}\frac{1}{\sqrt{n}e^{\sqrt{n}}}=0

so the divergence test is inconclusive. Our series isn't geometric, and it doesn't seem obvious how to compare it to a geometric series. However, the terms we're adding seem like they would make an integrable function.

Set f(x)=e−xxf(x)=\frac{e^{-\sqrt{x}}}{\sqrt{x}}. For x≥1x\ge1, this function is positive and decreasing (since it is the product of the two positive decreasing functions e−xe^{-\sqrt x} and 1x\frac1{\sqrt x}). We use the integral test with this function. Using the substitution u=xu=\sqrt x, so that du=12x dx\dee{u} = \frac1{2\sqrt x}\,\dee{x}, we see that

∫1∞f(x) dx=lim⁡R→∞∫1R(e−xx dx)=lim⁡R→∞(∫1Re−u⋅2 du)=lim⁡R→∞(−2e−u∣1R)=lim⁡R→∞(−2e−R+2e−1)=0+2e−1,\begin{align*} \int_1^\infty f(x)\,\dee{x} &=\lim_{R\rightarrow\infty}\int_1^R \bigg( \frac{e^{-\sqrt{x}}}{\sqrt{x}}\,\dee{x} \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( \int_1^{\sqrt{R}} e^{-u}\cdot2\,\dee{u} \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( {-}2e^{-u}\Big|_1^{\sqrt{R}} \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( {-}2e^{-\sqrt R} + 2e^{-\sqrt1} \bigg) = 0 + 2e^{-1}, \end{align*}

and so this improper integral converges. By the integral test, the given series also converges.

Q31Stage 2Past exam · 2016Q6

Use the comparison test (not the limit comparison test) to show whether the series
∑n=2∞3n2−7n3\displaystyle \sum_{n=2}^{\infty} \frac{\sqrt{3 n^2 - 7}}{n^{3}} converges or diverges.

Hint

Review Example 3.3.9 in the CLP-2 text for developing intuition about comparisons, and Example 3.3.10 for an example where finding an appropriate comparison series calls for some creativity.

Answer

The series ∑n=2∞3n2\displaystyle \sum_{n=2}^{\infty} \frac{\sqrt{3}}{n^2} converges by the pp–test with p=2p=2.

Note that

0<an=3n2−7n3<3n2n3=3n2\begin{align*} 0< a_n = \frac{\sqrt{3 n^2 - 7}}{n^3} < \frac{\sqrt{3n^2}}{n^3} = \frac{\sqrt{3}}{n^2} \end{align*}

for all n≥2n\ge 2. As the series ∑n=2∞3n2\sum\limits_{n=2}^\infty \frac{\sqrt3}{n^2} converges, the comparison test says that ∑n=2∞3n2−7n3\sum\limits_{n=2}^\infty \frac{\sqrt{3 n^2 - 7}}{n^{3}} converges too.

Full solution

We first develop some intuition. For very large nn, 3n23n^2 dominates 77 so that

3n2−7n3≈3n2n3=3n2\begin{align*} \frac{\sqrt{3 n^2 - 7}}{n^3} \approx \frac{\sqrt{3 n^2}}{n^3} =\frac{\sqrt{3}}{n^2} \end{align*}

The series ∑n=2∞1n2\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^2} converges by the pp–test with p=2p=2, so we expect the given series to converge too.

To verify that our intuition is correct, it suffices to observe that

0<an=3n2−7n3<3n2n3=3n2=cn\begin{align*} 0< a_n = \frac{\sqrt{3 n^2 - 7}}{n^3} < \frac{\sqrt{3n^2}}{n^3} = \frac{\sqrt{3}}{n^2} =c_n \end{align*}

for all n≥2n\ge 2. As the series ∑n=2∞cn\sum\limits_{n=2}^\infty c_n converges, the comparison test says that ∑n=2∞an\sum\limits_{n=2}^\infty a_n converges too.

Q32Stage 2Past exam · M105 2014A

Determine whether the series ∑k=1∞k4+13k5+9\displaystyle\sum_{k=1}^\infty\frac{ \sqrt[3]{k^4+1} } {\sqrt{k^5+9}} converges.

Hint

What does the summand look like when kk is very large?

Answer

The series converges.

Full solution

We first develop some intuition. For very large kk, k4k^4 dominates 11 so that the numerator k4+13≈k43=k4/3\sqrt[3]{k^4+1} \approx \sqrt[3]{k^4} = k^{4/3}, and k5k^5 dominates 9 so that the denominator k5+9≈k5=k5/2\sqrt{k^5+9} \approx \sqrt{k^5} = k^{5/2} and the summand

k4+13k5+9≈k4/3k5/2=1k7/6\begin{align*} \frac{ \sqrt[3]{k^4+1} }{\sqrt{k^5+9}} \approx \frac{k^{4/3}}{k^{5/2}} =\frac{1}{k^{7/6}} \end{align*}

The series ∑n=1∞1k7/6\displaystyle \sum_{n=1}^{\infty} \frac{1}{k^{7/6}} converges by the pp–test with p=76>1p=\frac{7}{6}>1, so we expect the given series to converge too.

To verify that our intuition is correct, we apply the limit comparison test with

ak=k4+13k5+9andbk=1k7/6=k4/3k5/2\begin{align*} a_k= \frac{ \sqrt[3]{k^4+1} }{\sqrt{k^5+9}} \quad\text{and}\quad b_k= \frac{1}{k^{7/6}}=\frac{k^{4/3}}{k^{5/2}} \end{align*}

which is valid since

lim⁡k→∞akbk=lim⁡k→∞k4+13/k4/3k5+9/k5/2=lim⁡k→∞1+1/k431+9/k5=1\begin{equation*} \lim_{k\rightarrow\infty} \frac{a_k}{b_k} =\lim_{k\rightarrow\infty}\frac{ \sqrt[3]{k^4+1}/k^{4/3}} {\sqrt{k^5+9}/k^{5/2}} =\lim_{k\rightarrow\infty}\frac{ \sqrt[3]{1+1/k^4} }{\sqrt{1+9/k^5}} =1 \end{equation*}

exists. Since the series ∑k=1∞bk\sum\limits_{k=1}^\infty b_k is a convergent pp–series (with ratio p=76>1p=\frac{7}{6}>1), the given series converges.

Note: to apply the direct comparison test with our chosen comparison series, we would need to show that

k4+13k5+9≤1k7/6\frac{\sqrt[3]{k^4+1}}{\sqrt{k^5+9}} \leq \frac{1}{k^{7/6}}

for all kk sufficiently large. However, this is not true: the opposite inequality holds when kk is large.

Q33Stage 2Past exam · 2016A

Does ∑n=1∞n42n/3(2n+7)4\displaystyle\sum_{n=1}^\infty\frac{n^4 2^{n/3}}{(2n+7)^4} converge or diverge?

Hint

What does the summand look like when nn is very large?

Answer

It diverges.

Full solution

  • Let's see whether the divergence test works here.

    lim⁡n→∞n42n/3(2n+7)4(1n41n4)=lim⁡n→∞2n/3(2+7/n)4=lim⁡n→∞2n/3(2+0)4=∞\lim_{n\to\infty} \frac{n^4 2^{n/3}}{(2n+7)^4}\left(\frac{\frac{1}{n^4}}{\frac{1}{n^4}}\right) = \lim_{n\to\infty} \frac{2^{n/3}}{(2+7/n)^4} = \lim_{n\to\infty} \frac{2^{n/3}}{(2+0)^4} = \infty

    The summands of our series do not converge to zero. By the divergence test, the series diverges.

  • Let's develop some intuition for a comparison. For very large nn, 2n2n dominates 77 so that

    n42n/3(2n+7)4≈n42n/3(2n)4=1162n/3\begin{align*} \frac{n^4 2^{n/3}}{(2n+7)^4} \approx \frac{n^4 2^{n/3}}{(2n)^4} =\frac{1}{16}2^{n/3} \end{align*}

    The series ∑n=1∞2n/3\displaystyle \sum_{n=1}^{\infty} 2^{n/3} is a geometric series with ratio r=21/3>1r=2^{1/3}>1 and so diverges. (It also fails the divergence test.) We expect the given series to diverge too.

    To verify that our intuition is correct, we apply the limit comparison test with

    an=n42n/3(2n+7)4andbn=2n/3\begin{align*} a_n= \frac{n^4 2^{n/3}}{(2n+7)^4} \quad\text{and}\quad b_n= 2^{n/3} \end{align*}

    which is valid since

    lim⁡n→∞anbn=lim⁡n→∞n4(2n+7)4=lim⁡n→∞1(2+7/n)4=124\begin{equation*} \lim_{n\rightarrow\infty} \frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{n^4}{(2n+7)^4} =\lim_{n\rightarrow\infty}\frac{1}{{(2+7/n)}^4} =\frac{1}{2^4} \end{equation*}

    exists and is nonzero. Since the series ∑n=1∞bn\sum\limits_{n=1}^\infty b_n is a divergent geometric series (with ratio r=21/3>1r=2^{1/3}>1), the given series diverges. (It is possible to use the plain comparison test as well. One needs to show something like an=n42n/3(2n+7)4≥n42n/3(2n+7n)4=194bna_n = \frac{n^4 2^{n/3}}{(2n+7)^4} \ge \frac{n^4 2^{n/3}}{(2n+7n)^4} = \frac{1}{9^4}b_n.)

  • Alternately, one can apply the ratio test:

    lim⁡n→∞∣an+1an∣=lim⁡n→∞∣(n+1)42(n+1)/3/(2(n+1)+7)4n42n/3/(2n+7)4∣=lim⁡n→∞(n+1)4(2n+7)4n4(2n+9)4 2(n+1)/32n/3=lim⁡n→∞(1+1/n)4(2+7/n)4(2+9/n)4⋅21/3=1⋅21/3>1.\begin{align*} \lim_{n\to\infty} \bigg| \frac{a_{n+1}}{a_n} \bigg| &= \lim_{n\to\infty} \bigg| \frac{(n+1)^42^{(n+1)/3}/(2(n+1)+7)^4}{n^42^{n/3}/(2n+7)^4} \bigg| \\ &= \lim_{n\to\infty} \frac{(n+1)^4(2n+7)^4}{n^4(2n+9)^4}\ \frac{2^{(n+1)/3}}{2^{n/3}} \\ &=\lim_{n\to\infty} \frac{(1+1/n)^4(2+7/n)^4}{(2+9/n)^4} \cdot2^{1/3} = 1\cdot2^{1/3} > 1. \end{align*}

    Since the ratio of consecutive terms is greater than one, by the ratio test, the series diverges.

Q34Stage 2Past exam · 2014A

Determine, with explanation, whether each of the following series converge or diverge.

  1. ∑n=1∞1n2+1\displaystyle\sum_{n=1}^\infty\frac{1}{\sqrt{n^2+1}}

  2. ∑n=1∞ncos⁡(nπ)2n\displaystyle\sum_{n=1}^\infty\frac{n\cos(n\pi)}{2^n}

Hint

cos⁡(nπ)\cos(n\pi) is a sneaky way to write (−1)n(-1)^n.

Answer

(a) diverges (b) converges

Full solution

(a) For large nn, n2≫1n^2\gg 1 and so n2+1≈n2=n\sqrt{n^2+1}\approx \sqrt{n^2}= n. This suggests that we apply the limit comparison test with an=1n2+1a_n=\frac{1}{\sqrt{n^2+1}} and bn=1nb_n=\frac{1}{n}. Since

lim⁡n→∞anbn=lim⁡n→∞1/n2+11/n=lim⁡n→∞11+1/n2=1\begin{equation*} \lim_{n\rightarrow\infty} \frac{a_n}{b_n} =\lim_{n\rightarrow\infty} \frac{1/\sqrt{n^2+1}}{1/n} =\lim_{n\rightarrow\infty} \frac{1}{\sqrt{1+1/n^2}} =1 \end{equation*}

and since ∑n=1∞1n\sum\limits_{n=1}^\infty\frac{1}{n} diverges, the given series diverges.

(b) Since cos⁡(nπ)=(−1)n\cos(n\pi)=(-1)^n, the given series converges by the alternating series test. To check that an=n2na_n=\frac{n}{2^n} decreases to 00 as nn tends to infinity, note that

an+1an=(n+1)2−(n+1)n2−n=(1+1n)12\begin{equation*} \frac{a_{n+1}}{a_n} =\frac{(n+1)2^{-(n+1)}}{n2^{-n}} =\Big(1+\frac{1}{n}\Big)\frac{1}{2} \end{equation*}

is smaller than 11 (so that an+1≤ana_{n+1}\le a_n) for all n≥1n\ge 1, and is smaller than 34\frac{3}{4} (so an+1≤34ana_{n+1}\le \frac{3}{4}a_n) for all n≥2n\ge 2.

Q35Stage 2Past exam · M105 2012A

Determine whether the series

∑k=1∞k4−2k3+2k5+k2+k\begin{equation*} \sum_{k=1}^\infty\frac{k^4-2k^3+2}{k^5+k^2+k} \end{equation*}

converges or diverges.

Hint

What is the behaviour for large kk?

Answer

The series diverges.

Full solution

For large kk, k4≫2k3−2k^4\gg 2k^3-2 and k5≫k2+kk^5\gg k^2+k so

k4−2k3+2k5+k2+k≈k4k5=1k.\frac{k^4-2k^3+2}{k^5+k^2+k}\approx \frac{k^4}{k^5} =\frac{1}{k}.

This suggests that we apply the limit comparison test with ak=k4−2k3+2k5+k2+ka_k=\frac{k^4-2k^3+2}{k^5+k^2+k} and bk=1kb_k=\frac{1}{k}. Since

lim⁡k→∞akbk=lim⁡k→∞k4−2k3+2k5+k2+k⋅k1=lim⁡k→∞k5−2k4+2kk5+k2+k=lim⁡k→∞1−2/k+2/k41+1/k3+1/k4=1\begin{align*} \lim_{k\rightarrow\infty} \frac{a_k}{b_k} &=\lim_{k\rightarrow\infty} \frac{k^4-2k^3+2}{k^5+k^2+k}\cdot \frac{k}{1} =\lim_{k\rightarrow\infty} \frac{k^5-2k^4+2k}{k^5+k^2+k} =\lim_{k\rightarrow\infty} \frac{1-2/k+2/k^4}{1+1/k^3+1/k^4} \\ &=1 \end{align*}

and since ∑k=1∞1k\sum\limits_{k=1}^\infty\frac{1}{k} diverges (by the pp–test with p=1p=1), the given series diverges.

Q36Stage 2Past exam · M105 2015A

Determine whether each of the following series converge or diverge.

  1. ∑n=2∞n2+n+1n5−n\displaystyle\sum_{n=2}^\infty\frac{n^2+n+1}{n^5-n}

  2. ∑m=1∞3m+sin⁡mm2\displaystyle\sum_{m=1}^\infty\frac{3m+\sin\sqrt{m}}{m^2}

Hint

When mm is large, 3m+sin⁡m≈3m3m+\sin\sqrt{m}\approx 3m.

Answer

(a) converges (b) diverges

Full solution

(a) For large nn, n2≫n+1n^2\gg n+1 and so the numerator n2+n+1≈n2n^2+n+1\approx n^2. For large nn, n5≫nn^5\gg n and so the denominator n5−n≈n5n^5-n\approx n^5. So, for large nn,

n2+n+1n5−n≈n2n5=1n3.\frac{n^2+n+1}{n^5-n}\approx \frac{n^2}{n^5} =\frac{1}{n^3}.

This suggests that we apply the limit comparison test with an=n2+n+1n5−na_n=\frac{n^2+n+1}{n^5-n} and bn=1n3b_n=\frac{1}{n^3}. Since

lim⁡n→∞anbn=lim⁡n→∞(n2+n+1)/(n5−n)1/n3=lim⁡n→∞n5+n4+n3n5−n=lim⁡n→∞1+1/n+1/n21−1/n4=1\begin{align*} \lim_{n\rightarrow\infty} \frac{a_n}{b_n} &=\lim_{n\rightarrow\infty} \frac{(n^2+n+1)/(n^5-n)}{1/n^3} =\lim_{n\rightarrow\infty} \frac{n^5+n^4+n^3}{n^5-n} =\lim_{n\rightarrow\infty} \frac{1+1/n+1/n^2}{1-1/n^4} \\ &=1 \end{align*}

exists and is nonzero, and since ∑n=1∞1n3\sum\limits_{n=1}^\infty\frac{1}{n^3} converges (by the pp–test with p=3>1p=3>1), the given series converges.

(b) For large mm, 3m≫∣sin⁡m∣3m\gg |\sin\sqrt{m}| and so

3m+sin⁡mm2≈3mm2=3m.\frac{3m+\sin\sqrt{m}}{m^2}\approx \frac{3m}{m^2} =\frac{3}{m}.

This suggests that we apply the limit comparison test with am=3m+sin⁡mm2a_m=\frac{3m+\sin\sqrt{m}}{m^2} and bm=1mb_m=\frac{1}{m}. (We could also use bm=3mb_m=\frac{3}{m}.) Since

lim⁡m→∞ambm=lim⁡m→∞(3m+sin⁡m)/m21/m=lim⁡m→∞3m+sin⁡mm=lim⁡m→∞3+sin⁡mm=3\begin{align*} \lim_{m\rightarrow\infty} \frac{a_m}{b_m} &=\lim_{m\rightarrow\infty} \frac{(3m+\sin\sqrt{m})/m^2}{1/m} =\lim_{m\rightarrow\infty} \frac{3m+\sin\sqrt{m}}{m} =\lim_{m\rightarrow\infty} 3+\frac{\sin\sqrt{m}}{m} \\ &=3 \end{align*}

exists and is nonzero, and since ∑m=1∞1m\sum\limits_{m=1}^\infty\frac{1}{m} diverges (by the pp–test with p=1p=1), the given series diverges.

Q37Stage 2

Evaluate the following series, or show that it diverges: ∑n=5∞1en\displaystyle\sum_{n=5}^\infty \frac{1}{e^n}.

Hint

This is a geometric series, but it doesn't start at n=0n=0.

Answer

1e5−e4\dfrac{1}{e^5-e^4}

Full solution
∑n=5∞1en=∑n=5∞(1e)n=∑n=0∞(1e)n−∑n=04(1e)n=11−1e−1−(1e)51−1e=(1e)51−1e=1e5(1−1e)=1e5−e4\begin{align*} \sum_{n=5}^\infty \frac{1}{e^n}&=\sum_{n=5}^\infty \left(\frac{1}{e}\right)^n\\ &=\sum_{n=0}^\infty \left(\frac{1}{e}\right)^n-\sum_{n=0}^4 \left(\frac{1}{e}\right)^n\\ &=\dfrac{1}{1-\frac{1}{e}}-\dfrac{1-\left(\frac{1}{e}\right)^{5}}{1-\frac{1}{e}}\\ &=\dfrac{\left(\frac{1}{e}\right)^5}{1-\frac{1}{e}} =\dfrac{1}{e^5\left(1-\frac{1}{e}\right)}\\ &=\dfrac{1}{e^5-e^4} \end{align*}
Q38Stage 2Past exam · 2016Q5

Determine whether the series ∑n=2∞67n\displaystyle\sum_{n=2}^\infty\frac{6}{7^n} is convergent or divergent. If it is convergent, find its value.

Hint

The series is geometric.

Answer

17\frac{1}{7}

Full solution

This is a geometric series.

∑n=2∞67n=∑n=0∞67n+2=∑n=0∞672⋅17n\begin{align*}\sum_{n=2}^\infty\frac{6}{7^n}&=\sum_{n=0}^\infty\frac{6}{7^{n+2}}= \sum_{n=0}^\infty\frac{6}{7^2}\cdot\frac{1}{7^n}\end{align*}

We use Equation 3.2.2 in the CLP-2 text with a=672a=\frac{6}{7^2} and r=17r=\frac{1}{7}.

=672⋅11−17=642=17\begin{align*}&=\frac{6}{7^2}\cdot\frac{1}{1-\frac17}=\frac{6}{42}=\frac17\end{align*}
Q39Stage 2Past exam · 2013A

Determine, with explanation, whether each of the following series converge or diverge.

  1. 1+13+15+17+19+⋯1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots.

  2. ∑n=1∞2n+122n+1{\displaystyle\sum_{n=1}^\infty \frac{2n+1}{2^{2n+1}}}

Hint

The first series can be written as ∑n=1∞12n−1\displaystyle\sum_{n=1}^\infty \frac{1}{2n-1} .

Answer

(a) diverges by limit comparison with the harmonic series

(b) converges by the ratio test

Full solution

(a)

  • The given series is

    1+13+15+17+19+⋯=∑n=1∞an with an=12n−1\begin{align*} 1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots =\sum_{n=1}^\infty a_n \text{ with }a_n=\frac{1}{2n-1} \end{align*}

    First we'll develop some intuition by observing that, for very large nn, an≈12na_n\approx \frac{1}{2n}. We know that the series ∑n=1∞1n\sum\limits_{n=1}^\infty\frac{1}{n} diverges by the pp–test with p=1p=1. So let's apply the limit comparison test with bn=1nb_n=\frac{1}{n}. Since

    lim⁡n→∞anbn=lim⁡n→∞n2n−1=lim⁡n→∞12−1n=12\begin{align*} \lim_{n\rightarrow\infty}\frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{n}{2n-1} =\lim_{n\rightarrow\infty}\frac{1}{2-\frac{1}{n}} =\frac{1}{2} \end{align*}

    the series ∑n=1∞an\sum\limits_{n=1}^\infty a_n converges if and only if the series ∑n=1∞bn\sum\limits_{n=1}^\infty b_n converges. So the given series diverges.

  • The series

    1+13+15+17+19+⋯≥12+14+16+18+110+⋯=12(1+12+13+14+15+⋯)\begin{align*} 1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots &\ge \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\cdots \\ &=\frac{1}{2}\Big(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots\Big) \end{align*}

    The series in the brackets is the harmonic series which we know diverges, by the pp–test with p=1p=1. So the series on the right hand side diverges. By the direct comparison test, the series on the left hand side diverges too.

(b) We'll use the ratio test with an=(2n+1)22n+1a_n=\dfrac{(2n+1)}{2^{2n+1}}. Since

an+1an=(2n+3)22n+322n+1(2n+1)=14(2n+3)(2n+1)=14(2+3/n)(2+1/n)→14<1 as n→∞\begin{align*} \frac{a_{n+1}}{a_n} &=\frac{(2n+3)}{2^{2n+3}}\frac{2^{2n+1}}{(2n+1)} =\frac{1}{4}\frac{(2n+3)}{(2n+1)} =\frac{1}{4}\frac{(2+3/n)}{(2+1/n)} \rightarrow \frac{1}{4}<1\text{ as }n\rightarrow\infty \end{align*}

the series converges.

Q40Stage 2Past exam · M105 2013A

Determine, with explanation, whether each of the following series converges or diverges.

  1. ∑k=2∞k3k2−k{\displaystyle \sum_{k=2}^\infty \frac{\sqrt[3]{k}}{k^2-k}}.

  2. ∑k=1∞k1010k(k!)2(2k)!{\displaystyle \sum_{k=1}^\infty \frac{k^{10}10^k(k!)^2}{(2k)!}}.

  3. ∑k=3∞1k(log⁡k)(log⁡log⁡k){\displaystyle \sum_{k=3}^\infty \frac{1}{k(\log k) (\log\log k)}}.

Answer

(a) Converges by the limit comparison test with b=1k5/3b=\frac{1}{k^{5/3}}.

(b) Diverges by the ratio test.

(c) Diverges by the integral test.

Full solution

(a) For very large kk, k≪k2k\ll k^2 so that

an=k3k2−k≈k3k2=1k5/3.a_n=\frac{\sqrt[3]{k}}{k^2-k}\approx \frac{\sqrt[3]{k}}{k^2} =\frac{1}{k^{5/3}}.

We apply the limit comparison test with bk=1k5/3b_k=\frac{1}{k^{5/3}}. Since

lim⁡k→∞akbk=lim⁡k→∞k3/(k2−k)1/k5/3=lim⁡k→∞k2k2−k=lim⁡k→∞11−1/k=1\begin{align*} \lim_{k\rightarrow\infty}\frac{a_k}{b_k} =\lim_{k\rightarrow\infty}\frac{\sqrt[3]{k}/(k^2-k)}{1/k^{5/3}} =\lim_{k\rightarrow\infty}\frac{k^2}{k^2-k} =\lim_{k\rightarrow\infty}\frac{1}{1-1/k} =1 \end{align*}

exists and is nonzero, and ∑k=1∞1k5/3\sum\limits_{k=1}^\infty\frac{1}{k^{5/3}} converges (by the pp–test with p=53>1p=\frac{5}{3}>1), the given series converges by the limit comparison test.

(b) The kthk^{\rm th} term in this series is ak=k1010k(k!)2(2k)!a_k= \frac{k^{10}10^k(k!)^2}{(2k)!}. Factorials often work well with the ratio test, because they simplify so nicely in quotients.

ak+1ak=(k+1)1010k+1((k+1)!)2(2k+2)!⋅(2k)!k1010k(k!)2=10(k+1k)10(k+1)2(2k+2)(2k+1)=10(1+1k)10(1+1/k)2(2+2/k)(2+1/k)\begin{align*} \frac{a_{k+1}}{a_k} &=\frac{(k+1)^{10}10^{k+1}((k+1)!)^2}{(2k+2)!}\cdot\frac{(2k)!}{k^{10}10^k(k!)^2} =10\Big(\frac{k+1}{k}\Big)^{10}\frac{(k+1)^2}{(2k+2)(2k+1)}\\ &=10\Big(1+\frac{1}{k}\Big)^{10}\frac{(1+1/k)^2}{(2+2/k)(2+1/k)} \end{align*}

As kk tends to ∞\infty, this converges to 10×1×12×2>110\times 1\times\frac{1}{2\times 2}>1. So the series diverges by the ratio test.

(c) We'll use the integal test. The kthk^{\rm th} term in the series is ak=1k(log⁡k)(log⁡log⁡k)=f(k)a_k=\frac{1}{k(\log k) (\log\log k)}=f(k) with f(x)=1x(log⁡x)(log⁡log⁡x)f(x)=\frac{1}{x(\log x) (\log\log x)}, which is continuous, positive and decreasing for x≥3x\ge 3.

∫3∞f(x) dx=∫3∞dxx(log⁡x)(log⁡log⁡x)=lim⁡R→∞∫3Rdxx(log⁡x)(log⁡log⁡x)=lim⁡R→∞∫log⁡3log⁡Rdyylog⁡ywith y=log⁡x, dy=dxx=lim⁡R→∞∫log⁡log⁡3log⁡log⁡Rdttwith t=log⁡y, dt=dyy=lim⁡R→∞[log⁡t]log⁡log⁡3log⁡log⁡R=∞\begin{alignat*}{3} \int_3^\infty f(x)\ \dee{x} &= \int_3^\infty \frac{\dee{x}}{x(\log x) (\log\log x)} =\lim_{R\rightarrow\infty} \int_3^R \frac{\dee{x}}{x(\log x) (\log\log x)} \\[0.1in] &=\lim_{R\rightarrow\infty} \int_{\log 3}^{\log R} \frac{\dee{y}}{y\log y} &\qquad\text{with }y=\log x,\ \dee{y}=\frac{\dee{x}}{x} \\[0.1in] &= \lim_{R\rightarrow\infty}\int_{\log \log 3}^{\log \log R} \frac{\dee{t}}{t} & \qquad\text{with }t=\log y,\ \dee{t}=\frac{\dee{y}}{y} \\[0.1in] &=\lim_{R \to \infty}\Big[\log t\Big]_{\log \log 3}^{\log\log R}=\infty \end{alignat*}

Since the integral is divergent, the series is divergent as well by the integral test.

Q41Stage 2Past exam · 2016Q6

Determine whether the series ∑n=1∞n3−42n5−6n\displaystyle\sum_{n=1}^\infty\frac{n^3-4}{2n^5-6n} is convergent or divergent.

Hint

What does the summand look like when nn is very large?

Answer

It converges.

Full solution

For large nn, the numerator n3−4≈n3n^3-4\approx n^3 and the denominator 2n5−6n≈2n52n^5-6n\approx 2n^5, so the nnth term is approximately n32n5=12n2\frac{n^3}{2n^5}=\frac{1}{2n^2}. So we apply the limit comparison test with an=n3−42n5−6na_n=\frac{n^3-4}{2n^5-6n} and bn=1n2b_n=\frac{1}{n^2}. Since

lim⁡n→∞anbn=lim⁡n→∞(n3−4)/(2n5−6n)1/n2=lim⁡n→∞1−4n32−6n4=12\begin{equation*} \lim_{n\rightarrow\infty}\frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{(n^3-4)/(2n^5-6n)}{1/n^2} =\lim_{n\rightarrow\infty}\frac{1-\frac{4}{n^3}}{2-\frac{6}{n^4}} =\frac12 \end{equation*}

exists and is nonzero, the given series ∑n=1∞an\sum\limits_{n=1}^\infty a_n converges if and only if the series ∑n=1∞bn\sum\limits_{n=1}^\infty b_n converges. Since the series ∑n=1∞bn=∑n=1∞1n2\sum\limits_{n=1}^\infty b_n =\sum\limits_{n=1}^\infty\frac{1}{n^2} is a convergent pp-series (with p=2p=2), both series converge.

Q42Stage 2Past exam · 2016Q6

What is the smallest value of NN such that the partial sum ∑n=1N(−1)nn⋅10n\displaystyle\sum_{n=1}^N\frac{(-1)^n}{n\cdot 10^n} approximates ∑n=1∞(−1)nn⋅10n\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{n\cdot 10^n} within an accuracy of 10−610^{-6}?

Hint

Review the alternating series test, which is given in Theorem 3.3.14 in the CLP-2 text.

Answer

N=5N=5

Full solution

By the alternating series test, the error introduced when we approximate the series ∑n=1∞(−1)nn⋅10n\displaystyle \sum_{n=1}^\infty\frac{(-1)^n}{n\cdot 10^n} by ∑n=1N(−1)nn⋅10n\displaystyle \sum_{n=1}^N\frac{(-1)^n}{n\cdot 10^n} is at most the magnitude of the first omitted term, 1(N+1)10(N+1)\displaystyle\frac1{(N+1) 10^{(N+1)}}. By trial and error, we find that this expression becomes smaller than 10−610^{-6} when N+1≥6N+1\ge 6. So the smallest allowable value is N=5N=5.

Q43Stage 2Past exam · 2016Q5

It is known that ∑n=1∞(−1)n−1n2=π212\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{12} (you don't have to show this). Find NN so that SNS_N, the NthN^{\rm th} partial sum of the series, satisfies ∣π212−SN∣≤10−6| \frac{\pi^2}{12} - S_N | \le 10^{-6}. Be sure to say why your method can be applied to this particular series.

Hint

Review the alternating series test, which is given in Theorem 3.3.14 in the CLP-2 text.

Answer

N≥999N\geq 999

Full solution

The sequence {1n2}\{\frac{1}{n^2}\} decreases to zero as nn increases to infinity. So, by the alternating series error bound, which is given in Theorem 3.3.14 in the CLP-2 text, π212−SN\frac{\pi^2}{12}-S_N lies between zero and the first omitted term, (−1)N(N+1)2\frac{(-1)^{N}}{(N+1)^2}. We therefore need 1(N+1)2≤10−6\frac{1}{(N+1)^2} \leq 10^{-6}, which is equivalent to N+1≥103N+1 \geq 10^3 and N≥999N\geq 999.

Q44Stage 2Past exam · 2015A

The series ∑n=1∞(−1)n+1(2n+1)2\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n+1)^2} converges to some number SS (you don't have to prove this). According to the Alternating Series Estimation Theorem, what is the smallest value of NN for which the NthN^{\rm th} partial sum of the series is at most 1100\frac1{100} away from SS? For this value of NN, write out the NthN^{\rm th} partial sum of the series.

Hint

Review the alternating series test, which is given in Theorem 3.3.14 in the

CLP-2 text.

Answer

We need N=4N=4 and then S4=132−152+172−192S_4= \frac{1}{3^2}-\frac{1}{5^2} +\frac{1}{7^2} -\frac{1}{9^2}

Full solution

The error introduced when we approximate SS by the NthN^{\rm th} partial sum SN=∑n=1N(−1)n+1(2n+1)2S_N=\sum_{n=1}^N \frac{(-1)^{n+1}}{(2n+1)^2} lies between 00 and the first term dropped, which is (−1)n+1(2n+1)2∣n=N+1=(−1)N+2(2N+3)2\frac{(-1)^{n+1}}{(2n+1)^2}\Big|_{n=N+1}=\frac{(-1)^{N+2}}{(2N+3)^2}. So we need the smallest positive integer NN obeying

1(2N+3)2≤1100(2N+3)2≥1002N+3≥10N≥72\begin{align*} \frac{1}{(2N+3)^2} &\le\frac{1}{100} \\ (2N+3)^2&\ge 100 \\ 2N+3&\ge 10 \\ N&\ge\frac{7}{2} \end{align*}

So we need N=4N=4 and then

S4=132−152+172−192\begin{equation*} S_4= \frac{1}{3^2}-\frac{1}{5^2} +\frac{1}{7^2} -\frac{1}{9^2} \end{equation*}

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q45Stage 3Past exam · M121 2014A

Determine, with explanation, whether the following series converge or diverge.

  1. ∑n=1∞nn9nn!\displaystyle\sum_{n=1}^\infty\frac{n^n}{9^n n!}

  2. ∑n=1∞1nlog⁡n\displaystyle\sum_{n=1}^\infty\frac{1}{n^{\log n}}

Answer

(a) converges (b) converges

Full solution

(a) There are plenty of powers/factorials. So let's try the ratio test with an=nn9nn!a_n= \frac{n^n}{9^n n!}.

lim⁡n→∞an+1an=lim⁡n→∞(n+1)n+19n+1(n+1)!9nn!nn=lim⁡n→∞(n+1)n+1nn 9 (n+1)=lim⁡n→∞(1+1/n)n9=e9\begin{align*} \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n} &=\lim_{n\rightarrow\infty}\frac{(n+1)^{n+1}}{9^{n+1}(n+1)!}\frac{9^nn!}{n^n} =\lim_{n\rightarrow\infty}\frac{(n+1)^{n+1}}{n^n\,9\,(n+1)} =\lim_{n\rightarrow\infty}\frac{(1+1/n)^n}{9} =\frac{e}{9} \end{align*}

Here we have used that lim⁡n→∞(1+1/n)n=e\lim\limits_{n\rightarrow\infty}(1+1/n)^n=e. See Example 3.7.20 in the CLP-1 text, with x=1nx=\frac{1}{n} and a=1a=1. As e<9e<9, our series converges.

(b) We know that the series ∑n=1∞1n2\sum_{n=1}^\infty \frac{1}{n^2} converges, by the pp–test with p=2p=2, and also that log⁡n≥2\log n \ge 2 for all n≥e2n\ge e^2. So let's use the limit comparison test with an=1nlog⁡na_n=\frac{1}{n^{\log n}} and bn=1n2b_n=\frac{1}{n^2}.

lim⁡n→∞anbn=lim⁡n→∞1nlog⁡n⋅n21=lim⁡n→∞1nlog⁡n−2=0\begin{equation*} \lim_{n\rightarrow\infty}\frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{1}{n^{\log n}}\cdot\frac{n^2}{1} =\lim_{n\rightarrow\infty}\frac{1}{n^{\log n-2}} =0 \end{equation*}

So our series converges, by the limit comparison test.

Q46Stage 3Past exam · 2013A

(a) Prove that ∫2∞x+sin⁡x1+x2 dx\displaystyle \int_2^\infty\frac{x+\sin x}{1+x^2}\ \dee{x} diverges.

(b) Explain why you cannot conclude that ∑n=1∞n+sin⁡n1+n2\displaystyle\sum\limits_{n=1}^\infty \frac{n+\sin n}{1+n^2} diverges from part (a) and the Integral Test.

(c) Determine, with explanation, whether ∑n=1∞n+sin⁡n1+n2\displaystyle\sum\limits_{n=1}^\infty \frac{n+\sin n}{1+n^2} converges or diverges.

Hint

For part (a), see Example 1.12.23 in the

CLP-2 text.

For part (b), review Theorem 3.3.5 in the

CLP-2 text.

For part (c), see Example 3.3.12 in the

CLP-2 text.

Answer

(a) See the solution.

(b) f(x)=x+sin⁡x1+x2f(x)=\dfrac{x+\sin x}{1+x^2} is not a decreasing function.

(c) See the solution.

Full solution

(a)

    • Our first task is to identify the potential sources of impropriety for this integral.

    • The domain of integration extends to +∞+\infty. On the domain of integration the denominator is never zero so the integrand is continuous. Thus the only problem is at +∞+\infty.

    • Our second task is to develop some intuition about the behavior of the integrand for very large xx. When xx is very large:

      • ∣sin⁡x∣≤1≪x|\sin x|\le 1 \ll x, so that the numerator x+sin⁡x≈xx+\sin x\approx x, and

      • 1≪x21 \ll x^2, so that denominator 1+x2≈x21+x^2\approx x^2, and

      • the integrand x+sin⁡x1+x2≈xx2=1x\displaystyle\frac{x+\sin x}{1+x^2} \approx \frac{x}{x^2} =\frac{1}{x}

    • Now, since ∫2∞dxx\displaystyle\int_2^\infty\frac{\dee{x}}{x} diverges, we would expect ∫2∞x+sin⁡x1+x2 dx\displaystyle\int_2^\infty\frac{x+\sin x}{1+x^2}\ \dee{x} to diverge too.

    • Our final task is to verify that our intuition is correct. To do so, we set

      f(x)=x+sin⁡x1+x2g(x)=1x\begin{align*} f(x) &= \frac{x+\sin x}{1+x^2} & g(x) &= \frac{1}{x} \end{align*}

      and compute

      lim⁡x→∞f(x)g(x)=lim⁡x→∞x+sin⁡x1+x2÷1x=lim⁡x→∞(1+sin⁡x/x)x(1/x2+1)x2×x=lim⁡x→∞1+sin⁡x/x1/x2+1=1\begin{align*} \lim_{x\rightarrow\infty}\frac{f(x)}{g(x)} &=\lim_{x\rightarrow\infty} \frac{x+\sin x}{1+x^2}\div\frac{1}{x} \\ &=\lim_{x\rightarrow\infty} \frac{(1+\sin x/x)x}{(1/x^2+1)x^2}\times x \\ &=\lim_{x\rightarrow\infty} \frac{1+\sin x/x}{1/x^2+1} \\ &=1 \end{align*}

    • Since ∫2∞g(x) dx=∫2∞dxx\displaystyle\int_2^\infty g(x)\ \dee{x} = \int_2^\infty\frac{\dee{x}}{x} diverges, by Example 1.12.8 in the

      CLP-2 text (To change the lower limit of integration from 11 to 22, just apply Theorem 1.12.20 in the

      CLP-2 text.), with p=1p=1, Theorem 1.12.22(b) in the

      CLP-2 text now tells us that ∫2∞f(x) dx=∫2∞x+sin⁡x1+x2 dx\displaystyle\int_2^\infty f(x)\ \dee{x} = \int_2^\infty\frac{x+\sin x}{1+x^2}\ \dee{x} diverges too.

  • Let's break up the integrand as x+sin⁡x1+x2=x1+x2+sin⁡x1+x2\dfrac{x+\sin x}{1+x^2} = \dfrac{x}{1+x^2}+\dfrac{\sin x}{1+x^2}. First, we consider the integral ∫2∞sin⁡x1+x2 dx\displaystyle \int_2^\infty\frac{\sin x}{1+x^2}\ \dee{x}.

    • ∣sin⁡x∣1+x2≤11+x2\displaystyle\frac{|\sin x|}{1+x^2}\le \frac{1}{1+x^2}, so if we can show ∫11+x2 dx\displaystyle\int \frac{1}{1+x^2}\,\dee{x} converges, we can conclude that ∫∣sin⁡x∣1+x2 dx\displaystyle\int\frac{|\sin x|}{1+x^2}\,\dee{x} converges as well by the comparison test.

    • ∫2∞11+x2 dx≤∫2∞1x2 dx\displaystyle\int_2^\infty\frac{1}{1+x^2}\ \dee{x}\le \int_2^\infty\frac{1}{x^2}\ \dee{x}

    •  ∫2∞1x2 dx\displaystyle\ \int_2^\infty\frac{1}{x^2}\ \dee{x} converges (by the pp–test with p=2p=2)

    • So the integral ∫2∞sin⁡x1+x2 dx\displaystyle \int_2^\infty\frac{\sin x}{1+x^2}\ \dee{x} converges by the comparison test, and hence

    • ∫2∞sin⁡x1+x2 dx\displaystyle \int_2^\infty\frac{\sin x}{1+x^2}\ \dee{x} converges as well.

    Therefore, ∫2∞x+sin⁡x1+x2 dx\displaystyle \int_2^\infty\frac{x+\sin x}{1+x^2}\ \dee{x} converges if and only if ∫2∞x1+x2 dx\displaystyle \int_2^\infty\frac{x}{1+x^2}\ \dee{x} converges. But

    ∫2∞x1+x2 dx=lim⁡r→∞∫2rx1+x2 dx=lim⁡r→∞[12log⁡(1+x2)]2r=∞\begin{align*} \int_2^\infty\frac{x}{1+x^2}\ \dee{x} =\lim_{r\rightarrow\infty}\int_2^r\frac{x}{1+x^2}\ \dee{x} =\lim_{r\rightarrow\infty}\Big[\half\log(1+x^2)\Big]_2^r=\infty \end{align*}

    diverges, so ∫2∞x+sin⁡x1+x2 dx\displaystyle\int_2^\infty\frac{x+\sin x}{1+x^2}\ \dee{x} diverges.

(b) The problem is that f(x)=x+sin⁡x1+x2f(x)=\dfrac{x+\sin x}{1+x^2} is not a decreasing function. To see this, compute the derivative:

f′(x)=(1+cos⁡x)(1+x2)−(x+sin⁡x)(2x)(1+x2)2=(cos⁡x−1)x2−2xsin⁡x+1+cos⁡x(1+x2)2\begin{align*} f'(x)=\frac{(1+\cos x)(1+x^2)-(x+\sin x)(2x)}{{(1+x^2)}^2} =\frac{(\cos x-1)x^2-2x\sin x +1+\cos x}{{(1+x^2)}^2} \end{align*}

If x=2mπx=2m\pi, the numerator is 0−0+1+1>00-0+1+1>0.

Therefore, the integral test does not apply.

(c)

  • Set an=n+sin⁡n1+n2a_n= \frac{n+\sin n}{1+n^2}. We first try to develop some intuition about the behaviour of ana_n for large nn and then we confirm that our intuition was correct.

    • Step 1: Develop intuition.
      When n≫1n\gg 1, the numerator n+sin⁡n≈nn+\sin n\approx n, and the denominator 1+n2≈n21+n^2\approx n^2 so that an≈nn2=1na_n\approx \frac{n}{n^2}=\frac{1}{n} and it looks like our series should diverge by the pp–test (Example 3.3.6 in the CLP-2 text) with p=1p=1.

    • Step 2: Verify intuition.
      To confirm our intuition we set bn=1nb_n=\frac{1}{n} and compute the limit

      lim⁡n→∞anbn=lim⁡n→∞n+sin⁡n1+n21n=lim⁡n→∞n[n+sin⁡n]1+n2=lim⁡n→∞1+sin⁡nn1n2+1=1\begin{equation*} \lim_{n\rightarrow\infty}\frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{ \frac{n+\sin n}{1+n^2} } {\frac{1}{n}} =\lim_{n\rightarrow\infty}\frac{n[n+\sin n]} {1+n^2} =\lim_{n\rightarrow\infty}\frac{1+\frac{\sin n}{n}} {\frac{1}{n^2}+1} =1 \end{equation*}

      We already know that the series ∑n=1∞bn=∑n=1∞1n\sum\limits_{n=1}^\infty b_n =\sum\limits_{n=1}^\infty\frac{1}{n} diverges by the pp–test with p=1p=1. So our series diverges by the limit comparison test, Theorem 3.3.11 in the CLP-2 text.

  • Since ∣sin⁡n1+n2∣≤1n2\big|\frac{\sin n}{1+n^2}\big|\le\frac{1}{n^2} and the series ∑n=1∞1n2\sum\limits_{n=1}^\infty \frac{1}{n^2} converges by the pp–test with p=2p=2, the series ∑n=1∞sin⁡n1+n2\sum\limits_{n=1}^\infty \frac{\sin n}{1+n^2} converges. Hence ∑n=1∞n+sin⁡n1+n2\sum\limits_{n=1}^\infty \frac{n+\sin n}{1+n^2} converges if and only if the series ∑n=1∞n1+n2\sum\limits_{n=1}^\infty \frac{n}{1+n^2} converges. Now f(x)=x1+x2f(x)=\frac{x}{1+x^2} is a continuous, positive, decreasing function on [1,∞)[1,\infty) since

    f′(x)=(1+x2)−x(2x)(1+x2)2=1−x2(1+x2)2\begin{align*} f'(x)=\frac{(1+x^2)-x(2x)}{{(1+x^2)}^2} =\frac{1-x^2}{{(1+x^2)}^2} \end{align*}

    is negative for all x>1x>1. We saw in part (a) that the integral ∫2∞x1+x2 dx\int_2^\infty\frac{x}{1+x^2}\ \dee{x} diverges. So the integral ∫1∞x1+x2 dx\int_1^\infty\frac{x}{1+x^2}\ \dee{x} diverges too and the sum ∑n=1∞n1+n2\sum\limits_{n=1}^\infty \frac{n}{1+n^2} diverges by the integral test. So ∑n=1∞n+sin⁡n1+n2\sum\limits_{n=1}^\infty \frac{n+\sin n}{1+n^2} diverges.

Q47Stage 3Past exam · M121 1999A

Show that ∑n=1∞e−nn\displaystyle\sum\limits_{n=1}^\infty\frac{e^{-\sqrt{n}}}{\sqrt{n}} converges and find an interval of length 0.050.05 or less that contains its exact value.

Hint

The truncation error arising from the approximation ∑n=1∞e−nn≈∑n=1Ne−nn\displaystyle\sum_{n=1}^\infty \frac{e^{-\sqrt{n}}}{\sqrt n} \approx \sum_{n=1}^N \frac{e^{-\sqrt{n}}}{\sqrt n} is precisely EN=∑n=N+1∞e−nnE_N = \displaystyle\sum_{n=N+1}^\infty \frac{e^{-\sqrt{n}}}{\sqrt n}. You'll want to find a bound on this sum using the integral test.

A key observation is that, since f(x)=e−xxf(x) = \dfrac{e^{-\sqrt{x}}}{\sqrt{x}} is decreasing, we can show that

e−nn≤∫n−1ne−xx dx\frac{e^{-\sqrt{n}}}{\sqrt{n}} \leq \int_{n-1}^n \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \dee{x}

for every n≥1n \ge 1.

Answer

The sum is between 0.9035 and 0.9535.

Full solution

Note that e−xx=1xex\dfrac{e^{-\sqrt{x}}}{\sqrt{x}}=\dfrac{1}{\sqrt{x}e^{\sqrt{x}}} decreases as xx increases. Hence, for every n≥1n\ge 1,

e−xx≥ennfor x in the interval [n−1,n]So,∫n−1ne−xx dx≥∫n−1ne−nn dx=[e−nnx]x=n−1x=n=e−nn\begin{align*} \frac{e^{-\sqrt{x}}}{\sqrt{x}} &\geq \frac{e^{\sqrt{n}}}{\sqrt{n}}&\text{for }x\text{ in the interval }[n-1,n]\text{}\\ \text{So,}\qquad\color{blue} \int_{n-1}^n \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \dee{x}&\geq\int_{n-1}^n \frac{e^{-\sqrt{n}}}{\sqrt{n}}\ \dee{x}\\ &=\left[\frac{e^{-\sqrt{n}}}{\sqrt{n}}x\right]_{x=n-1}^{x=n}\\ &=\color{red}\frac{e^{-\sqrt{n}}}{\sqrt{n}} \end{align*}

Then, for every N≥1N\ge 1,

EN=∑n=N+1∞e−nn≤∑n=N+1∞∫n−1ne−xx dx=∫NN+1e−xx dx+∫N+1N+2e−xx dx+⋯=∫N∞e−xx dx\begin{align*} E_N&=\sum_{n=N+1}^\infty\textcolor{red}{\frac{e^{-\sqrt{n}}}{\sqrt{n}}} \le \sum_{n=N+1}^\infty\textcolor{blue}{\int_{n-1}^n \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \dee{x}} \\&=\int_{N}^{N+1}\frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \dee{x} +\int_{N+1}^{N+2} \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \dee{x}+\cdots\\ &=\int_{N}^\infty \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \dee{x} \end{align*}

Substituting y=xy=\sqrt{x}, dy=12dxx\dee{y}=\half\frac{\dee{x}}{\sqrt{x}},

∫N∞e−xx dx=2∫N∞e−y dy=−2e−y∣N∞=2e−N\begin{align*} \int_{N}^\infty \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \dee{x} =2\int_{\sqrt{N}}^\infty e^{-y}\ \dee{y} =-2e^{-y}\Big|_{\sqrt{N}}^\infty =2e^{-\sqrt{N}} \end{align*}

This shows that ∑n=N+1∞e−nn\sum_{n=N+1}^\infty\frac{e^{-\sqrt{n}}}{\sqrt{n}} converges and is between 00 and 2e−N2e^{-\sqrt{N}}. Since E14=2e−14=0.047E_{14}=2e^{-\sqrt{14}}=0.047, we may truncate the series at n=14n=14.

∑n=1∞e−nn=∑n=114e−nn+E14=+0.3679+0.1719+0.1021+0.0677+0.0478=+0.0352+0.0268+0.0209+0.0166+0.0134=+0.0109+0.0090+0.0075+0.0063+E14=0.9042+E14\begin{align*} \sum_{n=1}^\infty\frac{e^{-\sqrt{n}}}{\sqrt{n}} &=\sum_{n=1}^{14}\frac{e^{-\sqrt{n}}}{\sqrt{n}}+E_{14}\\ &=\phantom{+}0.3679+0.1719+0.1021+0.0677+0.0478\\ &\phantom{=}+0.0352+0.0268+0.0209+0.0166+0.0134\\ &\phantom{=}+0.0109+0.0090+0.0075+0.0063+E_{14}\\ &=0.9042+E_{14} \end{align*}

The sum is between 0.9035 and 0.9535. (This even allows for a roundoff error of 0.000050.00005 in each term as we were calculating the partial sum.)

Q48Stage 3Past exam · M121 2000A

Suppose that the series ∑n=1∞an\displaystyle\sum\limits_{n=1}^\infty a_n converges and that 1>an≥01>a_n\ge 0 for all nn. Prove that the series ∑n=1∞an1−an\displaystyle\sum\limits_{n=1}^\infty \frac{a_n}{1-a_n} also converges.

Hint

What does the fact that the series ∑n=0∞an\sum\limits_{n=0}^{\infty}a_n converges guarantee about the behavior of ana_n for large nn?

Answer

Since lim⁡n→∞an=0\lim\limits_{n\to \infty} a_n=0, there must be some integer NN such that 12>an≥0\half>a_n\ge 0 for all n>Nn>N. Then, for n>Nn>N,

an1−an≤an1−1/2=2an\begin{align*}\frac{a_n}{1-a_n} &\leq \frac{a_n}{1-1/2}=2a_n\end{align*}

From the information in the problem statement, we know

∑n=N+1∞2an=2∑n=N+1∞anconverges. \begin{align*}\sum_{n=N+1}^\infty 2a_n&=2\sum_{n=N+1}^\infty a_n\qquad\text{converges. }\end{align*}

So, by the direct comparison test,

∑n=N+1∞an1−anconverges as well. \begin{align*}\sum_{n=N+1}^\infty\frac{a_n}{1-a_n}&\qquad\text{converges as well. }\end{align*}

Since the convergence of a series is not affected by its first NN terms, as long as NN is finite, we conclude

∑n=1∞an1−anconverges.\begin{align*}\sum_{n=1}^\infty\frac{a_n}{1-a_n}&\qquad\text{converges.}\end{align*}
Full solution

Let's get some intuition to guide us through a proof. Since ∑n=1∞an\sum\limits_{n=1}^\infty a_n, converges ana_n must converge to zero as n→∞n\rightarrow\infty. So, when nn is quite large, an1−an≈an1−0=an1\frac{a_n}{1-a_n} \approx \frac{a_n}{1-0}=\frac{a_n}{1}, and we know ∑an\sum a_n converges. So, we want to separate the “large" indices from a finite number of smaller ones.

Since lim⁡n→∞an=0\lim\limits_{n\to \infty} a_n=0, there must be (We could have chosen any positive number strictly less than 1, not only 12\frac12.) some integer NN such that 12>an≥0\half>a_n\ge 0 for all n>Nn>N. Then, for n>Nn>N,

an1−an≤an1−1/2=2an\begin{align*}\frac{a_n}{1-a_n} &\leq \frac{a_n}{1-1/2}=2a_n\end{align*}

From the information in the problem statement, we know

∑n=N+1∞2an=2∑n=N+1∞anconverges. \begin{align*}\sum_{n=N+1}^\infty 2a_n&=2\sum_{n=N+1}^\infty a_n\qquad\text{converges. }\end{align*}

So, by the direct comparison test,

∑n=N+1∞an1−anconverges as well. \begin{align*}\sum_{n=N+1}^\infty\frac{a_n}{1-a_n}&\qquad\text{converges as well. }\end{align*}

Since the convergence of a series is not affected by its first NN terms, as long as NN is finite, we conclude

∑n=1∞an1−anconverges.\begin{align*}\sum_{n=1}^\infty\frac{a_n}{1-a_n}&\qquad\text{converges.}\end{align*}
Q49Stage 3Past exam · M105 2015A

Suppose that the series ∑n=0∞(1−an)\sum\limits_{n=0}^{\infty}(1-a_n) converges, where an>0a_n>0 for n=0,1,2,3,⋯n=0,1,2,3,\cdots. Determine whether the series ∑n=0∞2nan\sum\limits_{n=0}^\infty 2^n a_n converges or diverges.

Hint

What does the fact that the series ∑n=0∞(1−an)\sum\limits_{n=0}^{\infty}(1-a_n) converges guarantee about the behavior of ana_n for large nn?

Answer

It diverges.

Full solution

By the divergence test, the fact that ∑n=0∞(1−an)\sum\limits_{n=0}^{\infty}(1-a_n) converges guarantees that lim⁡n→∞(1−an)=0\lim\limits_{n\rightarrow\infty}(1-a_n)=0, or equivalently, that lim⁡n→∞an=1\lim\limits_{n\rightarrow\infty}a_n=1. So, by the divergence test, a second time, the fact that

lim⁡n→∞2nan=+∞\begin{align*} \lim_{n\rightarrow\infty}2^n a_n = +\infty \end{align*}

guarantees that ∑n=0∞2nan\sum\limits_{n=0}^\infty 2^n a_n diverges too.

Q50Stage 3Past exam · M105 2014A

Assume that the series ∑n=1∞nan−2n+1n+1\displaystyle\sum_{n=1}^\infty\frac{na_n-2n+1}{n+1} converges, where an>0a_n > 0 for n=1,2,⋯n = 1, 2, \cdots. Is the following series

−log⁡a1+∑n=1∞log⁡(anan+1)\begin{align*} -\log a_1 + \sum_{n=1}^\infty \log\Big(\frac{a_n}{a_{n+1}}\Big) \end{align*}

convergent? If your answer is NO, justify your answer. If your answer is YES, evaluate the sum of the series −log⁡a1+∑n=1∞log⁡(anan+1)-\log a_1 + \sum\limits_{n=1}^\infty \log\big(\frac{a_n}{a_{n+1}}\big).

Hint

What does the fact that the series ∑n=1∞nan−2n+1n+1\displaystyle\sum_{n=1}^\infty\frac{na_n-2n+1}{n+1} converges guarantee about the behavior of ana_n for large nn?

Answer

It converges to −log⁡2=log⁡12-\log 2 =\log\frac{1}{2},

Full solution

By the divergence test, the fact that ∑n=1∞nan−2n+1n+1\displaystyle\sum_{n=1}^\infty\frac{na_n-2n+1}{n+1} converges guarantees that lim⁡n→∞nan−2n+1n+1=0\lim\limits_{n\rightarrow\infty}\dfrac{na_n-2n+1}{n+1}=0, or equivalently, that

0=lim⁡n→∞nn+1an−lim⁡n→∞2n−1n+1=lim⁡n→∞an−2  ⟺  lim⁡n→∞an=2\begin{align*} 0=\lim_{n\rightarrow\infty}\frac{n}{n+1}a_n -\lim_{n\rightarrow\infty}\frac{2n-1}{n+1} =\lim_{n\rightarrow\infty}a_n -2 \iff \lim_{n\rightarrow\infty}a_n = 2 \end{align*}

The series of interest can be written −log⁡a1+∑n=1∞[log⁡(an)−log⁡(an+1)]-\log a_1 + \sum\limits_{n=1}^\infty \big[\log(a_n)-\log(a_{n+1})\big] which looks like a telescoping series. So we'll compute the partial sum

SN=−log⁡a1+∑n=1N[log⁡(an)−log⁡(an+1)]=−log⁡a1+[log⁡(a1)−log⁡(a2)]+[log⁡(a2)−log⁡(a3)]+⋯+[log⁡(aN)−log⁡(aN+1)]=−log⁡(aN+1)\begin{align*} S_N&=-\log a_1 + \sum_{n=1}^N \big[\log(a_n)-\log(a_{n+1})\big] \\ &=-\log a_1 + \big[\log(a_1)-\log(a_2)\big] + \big[\log(a_2)-\log(a_3)\big] + \cdots + \big[\log(a_N)-\log(a_{N+1})\big]\\ &=-\log(a_{N+1}) \end{align*}

and then take the limit N→∞N\rightarrow\infty

−log⁡a1+∑n=1∞[log⁡(an)−log⁡(an+1)]=lim⁡N→∞SN=−lim⁡N→∞log⁡(aN+1)=−log⁡2=log⁡12\begin{align*} -\log a_1 + \sum_{n=1}^\infty \big[\log(a_n)-\log(a_{n+1})\big] =\lim_{N\rightarrow\infty} S_N =- \lim_{N\rightarrow\infty} \log(a_{N+1}) =-\log 2 =\log\frac{1}{2} \end{align*}
Q51Stage 3Past exam · M121 2002A

Prove that if an≥0a_n\ge 0 for all nn and if the series ∑n=1∞an\displaystyle\sum_{n=1}^\infty a_n converges, then the series ∑n=1∞an2\displaystyle\sum_{n=1}^\infty a^2_n also converges.

Hint

What does the fact that the series ∑n=1∞an\sum_{n=1}^\infty a_n converges guarantee about the behavior of ana_n for large nn? When is x2≤xx^2\le x?

Answer

See the solution.

Full solution

We are told that ∑n=1∞an\sum_{n=1}^\infty a_n converges. Thus we must have that lim⁡n→∞an=0\lim\limits_{n\rightarrow\infty}a_n=0. In particular, there is an index NN such that 0≤an≤10\le a_n\le 1 for all n≥Nn\ge N. Then:

0≤an2≤anfor n>N\begin{align*}0\leq a_n^2 & \leq a_n\qquad\text{for }n>N\end{align*}

By the direct comparison test,

∑n=N+1∞an2converges.\begin{align*}\sum_{n=N+1}^\infty a_n^2& \qquad\text{converges.}\end{align*}

Since convergence doesn't depend on the first NN terms of a series for any finite NN,

∑n=1∞an2converges as well.\begin{align*}\sum_{n=1}^\infty a_n^2& \qquad\text{converges as well.}\end{align*}

A number of phenomena roughly follow a distribution called Zipf's law. We discuss some of these in Questions 52 and 53.

Q52Stage 3

Suppose the frequency of word use in a language has the following pattern:

The nn-th most frequently used word accounts for αn\dfrac{\alpha}{n} percent of the total words used.

So, in a text of 100 words, we expect the most frequently used word to appear α\alpha times, while the second-most-frequently used word should appear about α2\frac{\alpha}{2} times, and so on.

If books written in this language use 20,00020,000 distinct words, then the most commonly used word accounts for roughly what percentage of total words used?

Hint

If we add together the frequencies of all the words, they should amount to 100%. We can approximate this sum using ideas from Example 3.3.4 in the CLP-2 text.

Answer

About 9% to 10%

Full solution

The most-commonly used word makes up α\alpha percent of all the words. So, we want to find α\alpha.

If we add together the frequencies of all the words, they should amount to 100%. That is,

∑n=120,000αn=100\sum_{n=1}^{20,000} \frac{\alpha}{n}=100

We can approximate the sum (with α\alpha left as a parameter) using the ideas behind the integral test. (See Example 3.3.4.)

Figure from prob_s3.3, line 2

Figure from prob_s3.3, line 2

As we see in the diagram above, ∑n=1Nαn\displaystyle\sum_{n=1}^N \frac{\alpha}{n} (which is the sum of the areas of the rectangles) is greater than ∫1N+1αxdx\displaystyle\int_1^{N+1}\frac{\alpha}{x}\dee{x} (the area under the curve). That is,

∫1N+1αxdx<∑n=1Nαn .\int_1^{N+1} \frac{\alpha}{x}\dee{x}< \sum_{n=1}^N \frac{\alpha}{n}\,.

Using the fact that our language's 20,000 words make up 100% of the words used, we can find a lower bound for α\alpha.

100=∑n=120,000αn>∫120,001αxdx=[αlog⁡(x)]120,001=αlog⁡(20,001)α<100log⁡(20,001)\begin{align*} 100&= \sum_{n=1}^{20,000} \frac{\alpha}{n} >\int_1^{20,001} \frac{\alpha}{x}\dee{x} =\Big[\alpha \log(x)\Big]_{1}^{20,001} =\alpha\log(20,001)\\ \alpha&<\frac{100}{\log(20,001)} \end{align*}

We can find an upper bound for α\alpha in a similar manner.

Figure from prob_s3.3, line 2

Figure from prob_s3.3, line 2

From the diagram, we see ∑n=2Nαn\displaystyle\sum_{n=2}^N \frac{\alpha}{n} (which is the sum of the areas of the rectangles, excluding the first) is less than ∫1Nαxdx\displaystyle\int_1^{N}\frac{\alpha}{x}\dee{x}. (The reason for excluding the first rectangle is to avoid comparing our series to an integral that diverges.) That is,

∑n=2Nαn<∫1Nαxdx .\sum_{n=2}^N \frac{\alpha}{n}< \int_1^{N} \frac{\alpha}{x}\dee{x}\,.

Therefore,

100=∑n=120,000αn=α+∑n=220,000αn<α+∫120,00αxdx=α+αlog⁡(20,000)=α[1+log⁡(20,000)]α>1001+log⁡(20,000)\begin{align*} 100&= \sum_{n=1}^{20,000} \frac{\alpha}{n}= \alpha+\sum_{n=2}^{20,000} \frac{\alpha}{n}\\ &<\alpha+\int_1^{20,00} \frac{\alpha}{x}\dee{x}=\alpha+\alpha\log(20,000)=\alpha\big[1+\log(20,000)\big]\\ \alpha&>\frac{100}{1+\log(20,000)} \end{align*}

Using a calculator, we see

9.17<α<10.019.17 < \alpha < 10.01

So, the most-commonly used word makes up about 9-10 percent of the total words.

Q53Stage 3

Suppose the sizes of cities in a country adhere to the following pattern: if the largest city has population α\alpha, then the nn-th largest city has population αn\frac{\alpha}{n}.

If the largest city in this country has 2 million people and the smallest city has 1 person, then the population of the entire country is ∑n=12×1062×106n\sum_{n=1}^{2 \times 10^6}\frac{2\times 10^6}{n}. (For many nn's in this sum 2×106n\frac{2\times 10^6}{n} is not an integer. Ignore that.) Evaluate this sum approximately, with an error of no more than 1 million people.

Hint

We are approximating a finite sum — not an infinite series. To get greater accuracy, use exact values for the first several terms in the sum, and use an integral to approximate the rest.

Answer

The total population is between 29,820,091 and 30,631,021 people.

Full solution

Generalizing our work in Question 52, we find the approximations:

∫ab+11xdx<∑n=ab1n<∫a−1b1xdx\int_a^{b+1}\frac{1}{x}\dee{x}<\sum_{n=a}^b \frac{1}{n} < \int_{a-1}^{b}\frac{1}{x}\dee{x}

when a≥2a \ge 2. The inequality ∫ab+11xdx<∑n=ab1n\int\limits_a^{b+1}\frac{1}{x}\dee{x}<\sum\limits_{n=a}^b \frac{1}{n} can be read off of the sketch

Figure from prob_s3.3, line 2

Figure from prob_s3.3, line 2

and the inequality ∑n=ab1n<∫a−1b1xdx\sum\limits_{n=a}^b \frac{1}{n}<\int\limits_{a-1}^{b}\frac{1}{x}\dee{x} can be read off of the sketch

Figure from prob_s3.3, line 2

Figure from prob_s3.3, line 2

We will evaluate the total population by writing

∑n=12×1062×106n=∑n=1a−12×106n+∑n=a2×1062×106n\begin{equation*} \sum_{n=1}^{2 \times 10^6}\frac{2\times 10^6}{n} =\sum_{n=1}^{a-1}\frac{2\times 10^6}{n} +\sum_{n=a}^{2 \times 10^6}\frac{2\times 10^6}{n} \end{equation*}

and applying the above integral approximations to the second sum. We want our error to be less than one million, so we need to choose a value of aa such that:

2×106∫a−12×1061xdx⏟upper bound−2×106∫a2×106+11xdx⏟lower bound<106∫a−12×1061xdx−∫a2×106+11xdx<12[log⁡(2×106)−log⁡(a−1)]−[log⁡(2×106+1)−log⁡(a)]<12[log⁡(2×106)−log⁡(2×106+1)]+[log⁡(a)−log⁡(a−1)]<12log⁡(2×1062×106+1)+log⁡(aa−1)<12\begin{align*} \underbrace{2\times 10^6\int_{a-1}^{2\times 10^6} \frac{1}{x}\dee{x}}_{\text{upper bound}} - \underbrace{2\times 10^6\int_{a}^{2\times 10^6+1} \frac{1}{x}\dee{x}}_{\text{lower bound}} &<10^6\\ \int_{a-1}^{2\times 10^6} \frac{1}{x}\dee{x} - \int_{a}^{2\times 10^6+1} \frac{1}{x}\dee{x} &<\frac{1}{2}\\ \left[\log\left(2\times 10^6 \right)-\log(a-1)\right]-\left[\log\left(2\times 10^6+1 \right)-\log(a)\right]&<\frac12\\ \left[\log\left(2\times 10^6 \right)-\log\left(2\times 10^6+1 \right)\right]+\left[ \log(a)-\log(a-1)\right]&<\frac12\\ \log\left(\frac{2\times 10^6}{2\times 10^6+1} \right)+ \log\left(\frac{a}{a-1} \right)&<\frac{1}{2} \end{align*}

The first term is extremely close to 0, so we ignore it.

log⁡(aa−1)<12aa−1<e1/2=ea<ae−ee<a(e−1)ee−1<a\begin{align*} \log\left(\frac{a}{a-1} \right)&<\frac{1}{2}\\ \frac{a}{a-1}&<e^{1/2}=\sqrt{e}\\ a&<a\sqrt{e}-\sqrt{e}\\ \sqrt{e}&<a(\sqrt{e}-1)\\ \frac{\sqrt{e}}{\sqrt{e}-1}&<a \end{align*}

Since ee−1≈2.5\dfrac{\sqrt{e}}{\sqrt{e}-1} \approx 2.5, we use a=3a=3. That is, we will approximate the value of ∑n=32×1061n\displaystyle\sum_{n=3}^{2\times 10^6}\frac{1}{n} using an integral. Then, we will use that approximation to estimate our total population.

∫32×106+11xdx<∑n=32×1061n<∫22×1061xdxlog⁡(2×106+1)−log⁡(3)<∑n=32×1061n<log⁡(2×106)−log⁡(2)1+12+log⁡(2×106+1)−log⁡(3)<∑n=12×1061n<1+12+log⁡(2×106)−log⁡(2)32+log⁡(2×106+13)<∑n=12×1061n<32+6log⁡(10)2×106(32+log⁡(23×106+13))<∑n=12×1062×106n<2×106(32+6log⁡(10))29,820,091<population<30,631,021\begin{alignat*}{5} \int_3^{2\times 10^6+1}\frac{1}{x}\dee{x}&<\sum_{n=3}^{2\times 10^6}\frac{1}{n} &&<\int_2^{2\times10^6}\frac{1}{x}\dee{x} \\ \log\left(2\times 10^6+1\right)-\log(3) &<\sum_{n=3}^{2\times 10^6}\frac{1}{n} &&<\log\left(2\times10^6\right)-\log(2) \\ \textcolor{red}{1+\frac12}+\log\left(2\times 10^6+1\right)-\log(3) &<\sum_{n=\textcolor{red}{1}}^{2\times 10^6}\frac{1}{n} &&<\textcolor{red}{1+\frac12}+\log\left(2\times10^6\right)-\log(2) \\ \frac32+ \log\left(\frac{2\times 10^6+1}{3}\right) &<\sum_{n=1}^{2\times 10^6}\frac{1}{n} &&<\frac32+6\log(10) \\ \textcolor{red}{2\times10^6}\Big(\frac32+ \log\left(\tfrac{2}{3}\times 10^6+\tfrac{1}{3}\right)\Big) &<\sum_{n=1}^{2\times 10^6}\frac{\textcolor{red}{2\times10^6}}{n} &&<\textcolor{red}{2\times10^6}\Big(\frac32+6\log\left(10\right)\Big) \\ 29,820,091&<\text{population}&&<30,631,021 \end{alignat*}

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.