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Mathematics LibreTexts

4.3E: Exercises

This page is a draft and is under active development. 

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Exercise 4.3E.1

For each of the following sequences, if the divergence test applies, either state that limnan does not exist or find limnan. If the divergence test does not apply, state why.

1) an=nn+2

2) an=n5n23

Answer

Solution: limnan=0. Divergence test does not apply.

3) an=n3n2+2n+1

4) an=(2n+1)(n1)(n+1)2

Answer

Solution: limnan=2. Series diverges.

5) an=(2n+1)2n(3n2+1)n

6) an=2n3n/2

Answer

Solution: limnan= (does not exist). Series diverges.

7) an=2n+3n10n/2

8) an=e2/n

Answer

Solution: limnan=1. Series diverges.

9) an=cosn

10) an=tann

Answer

Solution: limnan does not exist. Series diverges

11) an=1cos2(1/n)sin2(2/n)

12) an=(11n)2n)

Answer

Solution: limnan=1/e2. Series diverges

13) an=lnnn

14) an=(lnn)2n

Answer

Solution: limnan=0. Divergence test does not apply.

Exercise 4.3E.2

State whether the given p-series converges.

1) n=11n

2) n=11nn

Answer

Solution: Series converges, p>1

3) n=113n2

4) n=113n4

Answer

Solution: Series converges, p=4/3>1.

5) \displaystyle \sum_{n=1}^∞\frac{n^e}{n^π}

6) \displaystyle \sum_{n=1}^∞\frac{n^π}{n^{2e}}

Answer

Solution: Series converges, \displaystyle p=2e−π>1.

Exercise \PageIndex{3}

Use the integral test to determine whether the following sums converge.

1) \displaystyle \sum_{n=1}^∞\frac{1}{\sqrt{n+5}}

2) \displaystyle \sum_{n=1}^∞\frac{1}{\sqrt[3]{n+5}}

Answer

Solution: Series diverges by comparison with \displaystyle ∫^∞_1\frac{dx}{(x+5)^{1/3}}.

3) \displaystyle \sum_{n=2}^∞\frac{1}{nlnn}

4) \displaystyle \sum_{n=1}^∞\frac{n}{1+n^2}

Answer

Solution: Series diverges by comparison with \displaystyle ∫^∞_1\frac{x}{1+x^2}dx.

5) \displaystyle \sum_{n=1}^∞\frac{e^n}{1+e^{2n}}

6) \displaystyle \sum_{n=1}^∞\frac{2n}{1+n^4}

Answer

Solution: Series converges by comparison with \displaystyle ∫^∞_1\frac{2x}{1+x^4}dx.

7) \displaystyle \sum_{n=2}^∞\frac{1}{nln^2n}

Exercise \PageIndex{4}

Express the following sums as \displaystyle p-series and determine whether each converges.

1) \displaystyle \sum_{n=1}^∞2^{−lnn} (Hint: \displaystyle 2^{−lnn}=1/n^{ln2}.)

Answer

Solution: \displaystyle 2^{−lnn}=1/n^{ln2}. Since \displaystyle ln2<1, diverges by \displaystyle p-series.

2) \displaystyle \sum_{n=1}^∞3^{−lnn} (Hint: \displaystyle 3^{−lnn}=1/n^{ln3}.)

3) \displaystyle \sum_{n=1}^n2^{−2lnn}

Answer

Solution: \displaystyle 2^{−2lnn}=1/n^{2ln2}. Since \displaystyle 2ln2−1<1, diverges by \displaystyle p-series.

4) \displaystyle \sum_{n=1}^∞n3^{−2lnn}

Exercise \PageIndex{5}

Use the estimate \displaystyle R_N≤∫^∞_Nf(t)dt to find a bound for the remainder \displaystyle R_N=\sum_{n=1}^∞a_n−\sum_{n=1}^Na_n where \displaystyle a_n=f(n).

1) \displaystyle \sum_{n=1}^{1000}\frac{1}{n^2}

Answer

Solution: \displaystyle R_{1000}≤∫^∞_{1000}\frac{dt}{t^2}=−\frac{1}{t}∣^∞_{1000}=0.001

2) \displaystyle \sum_{n=1}^{1000}\frac{1}{n^3}

3) \displaystyle \sum_{n=1}^{1000}\frac{1}{1+n^2}

Answer

Solution: \displaystyle R_{1000}≤∫^∞_{1000}\frac{dt}{1+t^2}=tan^{−1}∞−tan^{−1}(1000)=π/2−tan^{−1}(1000)≈0.000999

4) \displaystyle \sum_{n=1}^{100}n/2^n

Exercise \PageIndex{6}

[T] Find the minimum value of \displaystyle N such that the remainder estimate \displaystyle ∫^∞_{N+1}f<R_N<∫^∞_Nf guarantees that \displaystyle \sum_{n=1}^Na_n estimates \displaystyle \sum_{n=1}^∞a_n, accurate to within the given error.

1) \displaystyle a_n=\frac{1}{n^2}, error \displaystyle <10^{−4}

Answer

Solution: \displaystyle R_N<∫^∞_N\frac{dx}{x^2}=1/N,N>10^4

2) \displaystyle a_n=\frac{1}{n^{1.1}}, error \displaystyle <10^{−4}

3) \displaystyle a_n=\frac{1}{n^{1.01}}, error \displaystyle <10^{−4}

Answer

Solution:\displaystyle R_N<∫^∞_N\frac{dx}{x^{1.01}}=100N^{−0.01},N>10^{600}

4) \displaystyle a_n=\frac{1}{nln^2n}, error \displaystyle <10^{−3}

5) \displaystyle a_n=\frac{1}{1+n^2}, error \displaystyle <10^{−3}

Answer

Solution: \displaystyle R_N<∫^∞_N\frac{dx}{1+x^2}=π/2−tan^{−1}(N),N>tan(π/2−10^{−3})≈1000

Exercise \PageIndex{7}

In the following exercises, find a value of \displaystyle N such that \displaystyle R_N is smaller than the desired error. Compute the corresponding sum \displaystyle \sum_{n=1}^Na_n and compare it to the given estimate of the infinite series.

1) \displaystyle a_n=\frac{1}{n^{11}}, error \displaystyle <10^{−4}, \sum_{n=1}^∞\frac{1}{n^{11}}=1.000494…

2) \displaystyle a_n=\frac{1}{e^n}, error \displaystyle <10^{−5}, \sum_{n=1}^∞\frac{1}{e^n}=\frac{1}{e−1}=0.581976…

Answer

Solution: \displaystyle R_N<∫^∞_N\frac{dx}{e^x}=e^{−N},N>5ln(10), okay if \displaystyle N=12;\sum_{n=1}^{12}e^{−n}=0.581973.... Estimate agrees with \displaystyle 1/(e−1) to five decimal places.

3) \displaystyle a_n=\frac{1}{e^{n^2}}, error \displaystyle <10^{−5}. \sum_{n=1}^∞n/e^{n^2}=0.40488139857…

4) \displaystyle a_n=1/n^4, error \displaystyle <10^{−4}, \sum_{n=1}^∞1/n^4=π^4/90=1.08232...

Answer

Solution: \displaystyle R_N<∫^∞_Ndx/x^4=4/N^3,N>(4.10^4)^{1/3}, okay if \displaystyle N=35; \displaystyle \sum_{n=1}^{35}1/n^4=1.08231…. Estimate agrees with the sum to four decimal places.

5) \displaystyle a_n=1/n^6, error \displaystyle <10^{−6}, \sum_{n=1}^∞1/n^4=π^6/945=1.01734306...,

6) Find the limit as \displaystyle n→∞ of \displaystyle \frac{1}{n}+\frac{1}{n+1}+⋯+\frac{1}{2n}. (Hint: Compare to \displaystyle ∫^{2n}_n\frac{1}{t}dt.)

Answer

Solution: \displaystyle ln(2)

7) Find the limit as \displaystyle n→∞ of \displaystyle \frac{1}{n}+\frac{1}{n+1}+⋯+\frac{1}{3n}

Exercise \PageIndex{8}

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.

1) In certain applications of probability, such as the so-called Watterson estimator for predicting mutation rates in population genetics, it is important to have an accurate estimate of the number \displaystyle H_k=(1+\frac{1}{2}+\frac{1}{3}+⋯+\frac{1}{k}). Recall that \displaystyle T_k=H_k−lnk is decreasing. Compute \displaystyle T=\lim_{k→∞}T_k to four decimal places. (Hint: \displaystyle \frac{1}{k+1}<∫^{k+1}_k\frac{1}{x}dx.)

Answer

Solution: \displaystyle T=0.5772...

2) [T] Complete sampling with replacement, sometimes called the coupon collector’s problem, is phrased as follows: Suppose you have \displaystyle N unique items in a bin. At each step, an item is chosen at random, identified, and put back in the bin. The problem asks what is the expected number of steps \displaystyle E(N) that it takes to draw each unique item at least once. It turns out that \displaystyle E(N)=N.H_N=N(1+\frac{1}{2}+\frac{1}{3}+⋯+\frac{1}{N}). Find \displaystyle E(N) for \displaystyle N=10,20, and \displaystyle 50.

3) [T] The simplest way to shuffle cards is to take the top card and insert it at a random place in the deck, called top random insertion, and then repeat. We will consider a deck to be randomly shuffled once enough top random insertions have been made that the card originally at the bottom has reached the top and then been randomly inserted. If the deck has \displaystyle n cards, then the probability that the insertion will be below the card initially at the bottom (call this card \displaystyle B) is \displaystyle 1/n. Thus the expected number of top random insertions before \displaystyle B is no longer at the bottom is \displaystyle n. Once one card is below \displaystyle B, there are two places below \displaystyle B and the probability that a randomly inserted card will fall below \displaystyle B is \displaystyle 2/n. The expected number of top random insertions before this happens is \displaystyle n/2. The two cards below \displaystyle B are now in random order. Continuing this way, find a formula for the expected number of top random insertions needed to consider the deck to be randomly shuffled.

Answer

Solution: The expected number of random insertions to get \displaystyle B to the top is \displaystyle n+n/2+n/3+⋯+n/(n−1). Then one more insertion puts \displaystyle B back in at random. Thus, the expected number of shuffles to randomize the deck is \displaystyle n(1+1/2+⋯+1/n).

4) Suppose a scooter can travel \displaystyle 100 km on a full tank of fuel. Assuming that fuel can be transferred from one scooter to another but can only be carried in the tank, present a procedure that will enable one of the scooters to travel \displaystyle 100H_N km, where \displaystyle H_N=1+1/2+⋯+1/N.

5) Show that for the remainder estimate to apply on \displaystyle [N,∞) it is sufficient that \displaystyle f(x) be decreasing on \displaystyle [N,∞), but \displaystyle f need not be decreasing on \displaystyle [1,∞).

Answer

Solution: Set \displaystyle b_n=a_{n+N} and \displaystyle g(t)=f(t+N) such that \displaystyle f is decreasing on \displaystyle [t,∞).

6) [T] Use the remainder estimate and integration by parts to approximate \displaystyle \sum_{n=1}^∞n/e^n within an error smaller than \displaystyle 0.0001.

7) Does \displaystyle \sum_{n=2}^∞\frac{1}{n(lnn)^p} converge if \displaystyle p is large enough? If so, for which \displaystyle p?

Answer

Solution: The series converges for \displaystyle p>1 by integral test using change of variable.

8) [T] Suppose a computer can sum one million terms per second of the divergent series \displaystyle \sum_{n=1}^N\frac{1}{n}. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed \displaystyle 100.

9) [T] A fast computer can sum one million terms per second of the divergent series \displaystyle \sum_{n=2}^N\frac{1}{nlnn}. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed \displaystyle 100.

Answer

Solution: \displaystyle N=e^{e^{100}}≈e^{10^{43}} terms are needed.


4.3E: Exercises is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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