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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
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
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
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
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
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
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
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
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
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
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
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
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
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
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

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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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

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
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
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
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
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
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
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

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
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

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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.