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Sequences and Series

3.1 Sequences

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

1Stage 1Conceptual

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

Q1Stage 1

Assuming the sequences continue as shown, estimate the limit of each sequence from its graph.

Figure from prob_s3.1, line 1

Figure from prob_s3.1, line 1

Figure from prob_s3.1, line 10

Figure from prob_s3.1, line 10

Figure from prob_s3.1, line 21

Figure from prob_s3.1, line 21

Hint

Not every limit exists.

Answer

(a) −2-2 (b) 0 (c) the limit does not exist

Full solution
Q2Stage 1

Suppose ana_n and bnb_n are sequences, and an=bna_n=b_n for all n≥100n \geq 100, but an≠bna_n \neq b_n for n<100n < 100.

True or false: lim⁡n→∞an=lim⁡n→∞bn\displaystyle\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n.

Hint

100 isn't all that big when you're contemplating infinity. (Neither is any other number.)

Answer

true

Full solution
Q3Stage 1

Let {an}n=1∞\{a_n\}_{n=1}^{\infty}, {bn}n=1∞\{b_n\}_{n=1}^{\infty}, and {cn}n=1∞\{c_n\}_{n=1}^{\infty}, be sequences with lim⁡n→∞an=A\lim\limits_{n \to \infty}a_n=A, lim⁡n→∞bn=B\lim\limits_{n \to \infty}b_n=B, and lim⁡n→∞cn=C\lim\limits_{n \to \infty}c_n=C. Assume AA, BB, and CC are nonzero real numbers.

Evaluate the limits of the following sequences.

  1. an−bncn\dfrac{a_n-b_n}{c_n}

  2. cnn\dfrac{c_n}{n}

  3. a2n+5bn\dfrac{a_{2n+5}}{b_n}

Hint

lim⁡n→∞a2n+5=lim⁡n→∞an\displaystyle\lim_{n \to \infty}a_{2n+5}=\lim_{n \to \infty}a_{n}

Answer

(a) A−BC\dfrac{A-B}{C} (b) 0 (c) AB\dfrac{A}{B}

Full solution
Q4Stage 1

Give an example of a sequence {an}n=1∞\{a_n\}_{n=1}^{\infty} with the following properties:

  • an>1000a_n>1000 for all n≤1000n \leq 1000,

  • an+1<ana_{n+1}<a_n for all nn, and

  • lim⁡n→∞an=−2\lim\limits_{n \to \infty} a_n = -2

Hint

The sequence might be defined by different functions when nn is large than when nn is small.

Answer

Two possible answers, of many:

  • an={3000−n if n≤1000−2+1n if n>1000a_n = \begin{cases} 3000-n & \text{ if }n \leq 1000\\ -2+\frac{1}{n} & \text{ if }n > 1000 \end{cases}

  • an=1,002,001n−2a_n=\dfrac{1,002,001}{n}-2

Full solution
Q5Stage 1

Give an example of a sequence {an}n=1∞\{a_n\}_{n=1}^{\infty} with the following properties:

  • an>0a_n>0 for all even nn,

  • an<0a_n<0 for all odd nn,

  • lim⁡n→∞an\lim\limits_{n \to \infty} a_n does not exist.

Hint

Recall (−1)n(-1)^n is positive when nn is even, and negative when nn is odd.

Answer

One possible answer is an=(−1)n={−1,1,−1,1,−1,1,−1,…}a_n=(-1)^{n} = \{-1, 1,-1,1,-1,1,-1,\ldots\}.
Another is an=n(−1)n={−1,2,−3,4,−5,6,−7,…}a_n=n(-1)^{n} = \{-1, 2,-3,4,-5,6,-7,\ldots\}.

Full solution
Q6Stage 1

Give an example of a sequence {an}n=1∞\{a_n\}_{n=1}^{\infty} with the following properties:

  • an>0a_n>0 for all even nn,

  • an<0a_n<0 for all odd nn,

  • lim⁡n→∞an\lim\limits_{n \to \infty} a_n exists.

Hint

Modify your answer from Question 5, but make the terms approach zero.

Answer

One sequence of many possible is an=(−1)nn={−1, 12, −13, 14, −15, 16, …}\displaystyle a_n = \frac{(-1)^n}{n} = \left\{-1,\ \frac12,\ -\frac13,\ \frac14,\ -\frac15,\ \frac16,\ \ldots \right\}.

Full solution
Q7Stage 1

The limits of the sequences below can be evaluated using the squeeze theorem. For each sequence, choose an upper bounding sequence and lower bounding sequence that will work with the squeeze theorem.

  1. an=sin⁡nna_n = \dfrac{\sin n}{n}

  2. bn=n2en(7+sin⁡n−5cos⁡n)b_n = \dfrac{n^2}{e^n(7+\sin n - 5\cos n)}

  3. cn=(−n)−nc_n = (-n)^{-n}

Hint

(−n)−n=(−1)nnn(-n)^{-n} = \dfrac{(-1)^n}{n^n}

Answer

Some possible answers:

  1. −1n≤sin⁡nn≤1n\dfrac{-1}{n}\le \dfrac{\sin n}{n}\le \dfrac{1}{n}

  2. n213en≤n2en(7+sin⁡n−5cos⁡n)≤n2en\dfrac{n^2}{13e^n} \le \dfrac{n^2}{e^n(7+\sin n - 5\cos n)} \le \dfrac{n^2}{e^n} or 0≤n2en(7+sin⁡n−5cos⁡n)≤n2en0\le \dfrac{n^2}{e^n(7+\sin n - 5\cos n)} \le \dfrac{n^2}{e^n}

  3. −1nn≤(−n)−n≤1nn\dfrac{-1}{n^n} \le (-n)^{-n} \le \dfrac{1}{n^n}

Full solution
Q8Stage 1

Below is a list of sequences, and a list of functions.

  1. Match each sequence {an}n=1∞\big\{a_n\big\}_{n=1}^\infty to any and all functions f(x)f(x) such that f(n)=anf(n)=a_n for all positive whole numbers nn.

  2. Match each sequence {an}n=1∞\big\{a_n\big\}_{n=1}^\infty to any and all functions f(x)f(x) such that lim⁡n→∞an=lim⁡x→∞f(x)\displaystyle\lim_{n \to \infty}a_n = \lim_{x \to \infty}f(x).

an=1+1nf(x)=cos⁡(πx)bn=1+1∣n∣g(x)=cos⁡(πx)xcn=e−nh(x)={x+1xx is a whole number1elsedn=(−1)ni(x)={x+1xx is a whole number0elseen=(−1)nnj(x)=1ex\begin{align*} a_n &= 1+\dfrac{1}{n} & f(x) &= \cos(\pi x)\\ b_n &= 1+\dfrac{1}{|n|} & g(x) &= \dfrac{\cos (\pi x)}{x}\\ c_n&=e^{-n} & h(x)&=\begin{cases}\frac{x+1}{x}&x\text{ is a whole number}\\ 1&\text{else} \end{cases}\\ d_n&=(-1)^n & i(x)&=\begin{cases}\frac{x+1}{x}&x\text{ is a whole number}\\ 0 & \text{else}\end{cases}\\ e_n&=\dfrac{(-1)^n}{n} & j(x)&=\dfrac{1}{e^x} \end{align*}
Hint

What might cause your answers in (a) and (b) to differ? Carefully read Theorem 3.1.6 in the CLP-2 text about convergent functions and their corresponding sequences.

Answer

(a) an=bn=h(n)=i(n)a_n=b_n=h(n)=i(n), cn=j(n)c_n = j(n), dn=f(n)d_n=f(n), en=g(n)e_n=g(n)
(b) lim⁡n→∞an=lim⁡n→∞bn=lim⁡x→∞h(x)=1\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} b_n = \lim\limits_{x \to \infty} h(x)=1, lim⁡n→∞cn=lim⁡n→∞en=lim⁡x→∞g(x)=lim⁡x→∞j(x)=0\lim\limits_{n \to \infty} c_n=\lim\limits_{n \to \infty} e_n=\lim\limits_{x \to \infty} g(x) = \lim\limits_{x \to \infty} j(x) =0,
lim⁡n→∞dn\lim\limits_{n \to \infty} d_n, lim⁡x→∞f(x)\lim\limits_{x\rightarrow\infty} f(x) and lim⁡x→∞i(x)\lim\limits_{x\rightarrow\infty} i(x) do not exist.

Full solution
Q9Stage 1

Let {an}n=1∞\{a_n\}_{n=1}^\infty be a sequence defined by an=cos⁡na_n = \cos n.

  1. Give three different whole numbers nn that are within 0.1 of an odd integer multiple of π\pi, and find the corresponding values of ana_n.

  2. Give three different whole numbers nn such that ana_n is close to 00. Justify your answers.

  3. Give three different whole numbers nn such that ana_n is close to 11. Justify your answers.

Remark: this demonstrates intuitively, though not rigorously, why lim⁡n→∞cos⁡n\lim\limits_{n \to \infty}\cos n is undefined. We consistently find terms in the sequence that are close to −1-1, and also consistently find terms in the sequence that are close to 0. Contrast this to a sequence like {cos⁡(2πn)}\big\{\cos(2\pi n)\big\}, whose terms are always 1, and whose limit therefore is 1. It is possible to turn the ideas of this question into a rigorous proof that lim⁡n→∞cos⁡n\lim\limits_{n \to \infty}\cos n is undefined. See the solution.

Hint

You can use the fact that π\pi is somewhat close to 227\dfrac{22}{7}, or you can use trial and error.

Answer

(a) Some possible answers: a22≈−0.99996a_{22}\approx -0.99996, a66≈−0.99965a_{66}\approx -0.99965, and a110≈−0.99902a_{110}\approx -0.99902.

(b) Some possible answers: a11≈0.0044a_{11}\approx 0.0044, a33≈−0.0133a_{33}\approx -0.0133, and a55≈0.0221a_{55}\approx 0.0221.
The integers 11, 33, and 55 were found by approximating π\pi by 227\dfrac{22}{7} and finding when an odd multiple of 117\dfrac{11}{7} (which is the corresponding approximation of π2\dfrac{\pi}{2}) is an integer.

(c) Some possible answers: a44≈0.9998a_{44}\approx 0.9998, a132≈0.9986a_{132}\approx 0.9986 and a220≈0.09961a_{220}\approx 0.09961.
See the solution for how we found them.

Full solution

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q10Stage 2

Determine the limits of the following sequences.

  1. an=3n2−2n+54n+3a_n = \dfrac{3n^2-2n+5}{4n+3}

  2. bn=3n2−2n+54n2+3b_n = \dfrac{3n^2-2n+5}{4n^2+3}

  3. cn=3n2−2n+54n3+3c_n = \dfrac{3n^2-2n+5}{4n^3+3}

Hint

You can compare the leading terms, or factor a high power of nn from the numerator and denominator.

Answer

(a) ∞\infty (b) 34\dfrac{3}{4} (c) 0

Full solution
Q11Stage 2

Determine the limit of the sequence an=4n3−21ne+1na_n = \dfrac{4n^3-21}{n^e+\frac{1}{n}}.

Hint

This isn't a rational expression, but you can treat it in a similar way. Recall e<3e<3.

Answer

∞\infty

Full solution
Q12Stage 2

Determine the limit of the sequence bn=n4+19n+3b_n = \dfrac{\sqrt[4]{n}+1}{\sqrt{9n+3}}.

Hint

The techniques of evaluating limits of rational sequences are again useful here.

Answer

0

Full solution
Q13Stage 2

Determine the limit of the sequence cn=cos⁡(n+n2)nc_n = \dfrac{\cos(n+n^2)}{n}.

Hint

Use the squeeze theorem.

Answer

0

Full solution
Q14Stage 2

Determine the limit of the sequence an=nsin⁡nn2a_n = \dfrac{n^{\sin n}}{n^2}.

Hint

1n≤nsin⁡n≤n\displaystyle\frac{1}{n}\leq n^{\sin n}\leq n

Answer

0

Full solution
Q15Stage 2

Determine the limit of the sequence dn=e−1/nd_n = e^{-1/n}.

Hint

e−1/n=1e1/ne^{-1/n} = \dfrac{1}{e^{1/n}}; what happens to 1n\dfrac{1}{n} as nn grows?

Answer

1

Full solution
Q16Stage 2

Determine the limit of the sequence an=1+3sin⁡(n2)−2sin⁡nna_n = \dfrac{1+3\sin(n^2)-2\sin n}{n}.

Hint

Use the squeeze theorem.

Answer

0

Full solution
Q17Stage 2

Determine the limit of the sequence bn=en2n+n2b_n=\dfrac{e^n}{2^n+n^2}.

Hint

L'H^opital's rule might help you decide what happens if you are unsure.

Answer

∞\infty

Full solution
Q18Stage 2Past exam · M105 2012A

Find the limit, if it exists, of the sequence {ak}\big\{a_k\big\}, where

ak=k!sin⁡3k(k+1)!\begin{equation*} a_k=\frac{k!\sin^3 k}{(k+1)!} \end{equation*}
Hint

Simplify aka_k.

Answer

lim⁡k→∞ak=0\lim\limits_{k\rightarrow\infty}a_k= 0.

Full solution
Q19Stage 2Past exam · 2013A

Consider the sequence {(−1)nsin⁡(1n)}\Big\{(-1)^n\sin\big(\frac{1}{n}\big)\Big\}. State whether this sequence converges or diverges, and if it converges give its limit.

Hint

What happens to 1n\dfrac{1}{n} as nn gets very big?

Answer

The sequence converges to 00.

Full solution
Q20Stage 2Past exam · 2016Q5

Evaluate lim⁡n→∞[6n2+5nn2+1+3cos⁡(1/n2)]\displaystyle\lim_{n\rightarrow\infty}\left[\frac{6n^2+5n}{n^2+1} +3\cos(1/n^2) \right].

Hint

cos⁡0=1\cos 0 =1

Answer

99

Full solution

3Stage 3Application

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

Q21Stage 3Past exam · M105 2015A

Find the limit of the sequence {log⁡(sin⁡1n)+log⁡(2n)}\displaystyle\left\{\log\left(\sin\frac{1}{n}\right)+\log(2n)\right\}.

Hint

This is trickier than it looks. Write 1n=x\dfrac{1}{n}=x and look at the limit as x→0x\rightarrow 0.

Answer

log⁡2\log 2

Full solution
Q22Stage 3

Evaluate lim⁡n→∞[n2+5n−n2−5n]\displaystyle\lim_{n \to \infty}\left[\sqrt{n^2+5n}-\sqrt{n^2-5n}\right].

Hint

Multiply and divide by the conjugate.

Answer

5

Full solution
Q23Stage 3

Evaluate lim⁡n→∞[n2+5n−2n2−5]\displaystyle\lim_{n \to \infty}\left[\sqrt{n^2+5n}-\sqrt{2n^2-5}\right].

Hint

Compared to Question 22, there's an easier path.

Answer

−∞-\infty

Full solution
Q24Stage 3

Evaluate the limit of the sequence {n[(2+1n)100−2100]}n=1∞\left\{n\left[\left(2+\frac1n\right)^{100}-2^{100}\right]\right\}_{n=1}^{\infty}.

Hint

Consider f′(x)f'(x), when f(x)=x100f(x)=x^{100}.

Answer

100⋅299100\cdot 2^{99}.

Full solution
Q25Stage 3

Write a sequence {an}n=1∞\{a_n\}_{n=1}^\infty whose limit is f′(a)f'(a) for a function f(x)f(x) that is differentiable at the point aa.

Your answer will depend on ff and aa.

Hint

Look to Question 24 for inspiration.

Answer

Possible answers are {an}={n[f(a+1n)−f(a)]}\{a_n\}=\left\{\displaystyle n\left[f\left(a+\frac{1}{n}\right)-f(a)\right]\right\}
or {an}={n[f(a)−f(a−1n)]}\{a_n\}=\left\{\displaystyle n\left[f(a)-f\left(a-\frac{1}{n}\right)\right]\right\}.

Full solution
Q26Stage 3

Let {An}n=3∞\{A_n\}_{n=3}^\infty be the area of a regular polygon with nn sides, with the distance from the centroid of the polygon to each corner equal to 1.

Figure from prob_s3.1, line 2

Figure from prob_s3.1, line 2

  1. By dividing the polygon into nn triangles, give a formula for AnA_n.

  2. What is lim⁡n→∞An\lim\limits_{n \to \infty} A_n?

Hint

The area of an isosceles triangle with two sides of length 1, meeting at an angle θ\theta, is 12sin⁡θ\frac{1}{2}\sin\theta.

Figure from prob_s3.1, line 2

Figure from prob_s3.1, line 2

Answer

(a) An=n2sin⁡(2πn)A_n = \dfrac{n}{2}\sin\left(\dfrac{2\pi}{n}\right) (b) π\pi

Full solution
Q27Stage 3

Suppose we define a sequence {fn}\{f_n\}, which depends on some constant xx, as the following:

fn(x)={1n≤x<n+10elsef_n(x) = \begin{cases} 1 & n \leq x < n+1\\ 0 & \text{else} \end{cases}

For a fixed constant x≥1x \ge 1, {fn}\{f_n\} is the sequence {0,0,0,…,0,1,0,…,0,0,0,…}\{0,0,0,\ldots,0,1,0,\ldots,0,0,0,\ldots\}. The sole nonzero element comes in position kk, where kk is what we get when we round xx down to a whole number. If x<1x<1, then the sequence consists of all zeroes.

Since we can plug in different values of xx, we can think of fn(x)f_n(x) as a function of sequences: a different xx gives you a different sequence. On the other hand, if we imagine fixing nn, then fn(x)f_n(x) is just a function, where fn(x)f_n(x) gives the nnth term in the sequence corresponding to xx.

  1. Sketch the curve y=f2(x)y=f_2(x).

  2. Sketch the curve y=f3(x)y=f_3(x).

  3. Define An=∫0∞fn(x) dxA_n = \int_0^\infty f_n(x)\,\dee{x}. Give a simple description of the sequence {An}n=1∞\{A_n\}_{n=1}^\infty.

  4. Evaluate lim⁡n→∞An\displaystyle\lim_{n \to \infty}A_n.

  5. Evaluate lim⁡n→∞fn(x)\displaystyle\lim_{n \to \infty} f_n(x) for a constant xx, and call the result g(x)g(x).

  6. Evaluate ∫0∞g(x) dx\displaystyle \int_0^\infty g(x)\,\dee{x}.

Hint

Every term of AnA_n is the same, and g(x)g(x) is a constant function.

Answer
  1. Figure from prob_s3.1, line 2

    Figure from prob_s3.1, line 2

  2. Figure from prob_s3.1, line 2

    Figure from prob_s3.1, line 2

  3. An=1A_n=1 for all nn

  4. lim⁡n→∞An=1\displaystyle\lim_{n \to \infty}A_n=1.

  5. g(x)=0g(x)=0

  6. ∫0∞g(x) dx=0\displaystyle \int_0^\infty g(x)\,\dee{x} =0.

Full solution
Q28Stage 3

Determine the limit of the sequence bn=(1+3n+5n2)n\displaystyle b_n=\left(1+\frac{3}{n}+\frac{5}{n^2}\right)^n.

Hint

You'll need to use a logarithm before you can apply l'H^opital's rule.

Answer

e3e^3

Full solution
Q29Stage 3

A sequence {an}n=1∞\big\{a_n\big\}_{n=1}^\infty of real numbers satisfies the recursion relation an+1=an+83a_{n+1} = \dfrac{a_n+8}{3} for n≥1n\ge 1.

  1. Suppose a1=4a_1=4. What is lim⁡n→∞an\lim\limits_{n \to \infty}a_n?

  2. Find xx if x=x+83x=\dfrac{x+8}{3}.

  3. Suppose a1=1a_1=1. Show that lim⁡n→∞an=L\lim\limits_{n\rightarrow\infty} a_n = L, where LL is the solution to equation above.

Hint

(a) Write out the first few terms of the sequence.
(c) Consider how an+1−La_{n+1}-L relates to an−La_{n}-L. What should happen to these numbers if ana_n converges to LL?

Answer

(a) 4 (b) x=4x=4 (c) see solution

Full solution
Q30Stage 3

Zipf's Law applied to word frequency can be phrased as follows:

The most-used word in a language is used nn times as frequently as the nn-th most word used in a language.

  1. Suppose the sequence {w1,w2,w3,…}\{w_1,w_2,w_3,\ldots\} is a list of all words in a language, where wnw_n is the word that is the nnth most frequently used. Let fnf_n be the frequency of word wnw_n. Is {f1,f2,f3,…}\{f_1,f_2,f_3,\ldots\} an increasing sequence or a decreasing sequence?

  2. Give a general formula for fnf_n, treating f1f_1 as a constant.

  3. Suppose in a language, w1w_1 (the most frequently used word) has frequency 6%. If the language follows Zipf's Law, then what frequency does w3w_3 have?

  4. Suppose f6=0.3%f_6=0.3\% for a language following Zipf's law. What is f10f_{10}?

  5. The word “the" is the most-used word in contemporary American English. In a collection of about 450 million words, “the" appeared 22,038,615 times. The second-most used word is “be," followed by “and." About how many usages of these words do you expect in the same collection of 450 million words?

Hint

Your answer from (b) will help you a lot with the subsequent parts.

Answer

(a) decreasing (b)fn=1nf1f_n=\frac{1}{n}f_1 (c) 2%2\% (d) 0.18%0.18\%
(e) “be": 11,019,308; “and": 7,346,205

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.