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9.3E: Exercises for Divergence and Integral Tests

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Divergence Test Problems

Consider the sequence for each series in exercises 1 - 14, 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) n=1nn+2

2) n=1n5n23

Answer
limnan=0. The Divergence Test does not apply.

3) n=1n3n2+2n+1

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

Answer
limnan=2. So the series diverges by the nth-Term Test for Divergence.

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

6) n=12n3n/2

Answer
limnan= (does not exist). So the series diverges by the nth-Term Test for Divergence.

7) n=12n+3n10n/2

8) n=1e2/n

Answer
limnan=1. So the series diverges by the nth-Term Test for Divergence.

9) n=1cosn

10) n=1tann

Answer
limnan does not exist. So the series diverges by the nth-Term Test for Divergence.

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

12) n=1(11n)2n

Answer
limnan=1/e2. So the series diverges by the nth-Term Test for Divergence.

13) n=1lnnn

14) n=1(lnn)2n

Answer
limnan=0. The Divergence Test does not apply.

p-Series Problems & Integral Test Problems

In exercises 15 - 20, state whether the given p-series converges.

15) n=11n

16) n=11nn

Answer
The series converges, since p=3/2>1.

17) n=113n2

18) n=113n4

Answer
The series converges, since p=4/3>1.

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

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

Answer
The series converges, since p=2e−π>1.

In exercises 21 - 27, use the integral test to determine whether the following sums converge.

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

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

Answer
The series diverges by the Integral Test since \displaystyle ∫^∞_1\frac{dx}{(x+5)^{1/3}} can be shown to diverge.

23) \displaystyle \sum_{n=2}^∞\frac{1}{n\ln n}

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

Answer
The series diverges by the Integral Test since \displaystyle ∫^∞_1\frac{x}{1+x^2}\,dx can be shown to diverge.

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

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

Answer
The series converges by the Integral Test since \displaystyle ∫^∞_1\frac{2x}{1+x^4}\,dx converges.

27) \displaystyle \sum_{n=2}^∞\frac{1}{n\ln^2n}

Express the sums in exercises 28 - 31 as p-series and determine whether each converges.

28) \displaystyle \sum_{n=1}^∞2^{−\ln n} \quad\Big(Hint: 2^{−\ln n}=\dfrac{1}{n^{\ln 2}}.\Big)

Answer
2^{−\ln n}=1/n^{\ln 2}. Since p=\ln 2<1, this series diverges by the p-series test.

29) \displaystyle \sum_{n=1}^∞3^{−\ln n} \quad\Big(Hint: 3^{−\ln n}=\dfrac{1}{n^{\ln 3}}.\Big)

30) \displaystyle \sum_{n=1}^n2^{−2\ln n}

Answer
2^{−2\ln n}=1/n^{2\ln 2}. Since p = 2\ln 2−1<1, this series diverges by the p-series test.

31) \displaystyle \sum_{n=1}^∞n3^{−2\ln n}

In exercises 32 - 35, 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 a_n=f(n).

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

Answer
\displaystyle R_{1000}≤∫^∞_{1000}\frac{dt}{t^2}=\lim_{b\to ∞}−\frac{1}{t}\bigg|^b_{1000}=\lim_{b\to ∞}\left(−\frac{1}{b}+\frac{1}{1000}\right)=0.001

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

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

Answer
\displaystyle R_{1000}≤∫^∞_{1000}\frac{dt}{1+t^2}=\lim_{b\to ∞} \left(\tan^{−1}b−\tan^{−1}(1000)\right)=π/2−\tan^{−1}(1000)≈0.000999

35) \displaystyle \sum_{n=1}^{100}\frac{n}{2^n}

[T] In exercises 36 - 40, find the minimum value of N such that the remainder estimate \displaystyle ∫^∞_{N+1}f(x)\,dx<R_N<∫^∞_N f(x)\,dx guarantees that \displaystyle \sum_{n=1}^Na_n estimates \displaystyle \sum_{n=1}^∞a_n, accurate to within the given error.

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

Answer
\displaystyle R_N<∫^∞_N\frac{dx}{x^2}=1/N,\;\text{for}\;N>10^4

37) a_n=\dfrac{1}{n^{1.1}}, error <10^{−4}

38) a_n=\dfrac{1}{n^{1.01}}, error <10^{−4}

Answer
\displaystyle R_N<∫^∞_N\frac{dx}{x^{1.01}}=100N^{−0.01},\;\text{for}\;N>10^{600}

39) a_n=\dfrac{1}{n\ln^2n}, error <10^{−3}

40) a_n=\dfrac{1}{1+n^2}, error <10^{−3}

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

In exercises 41 - 45, find a value of N such that 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.

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

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

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

43) a_n=\dfrac{1}{e^{n^2}}, error \displaystyle <10^{−5}, \sum_{n=1}^∞\dfrac{1}{e^{n^2}}=0.40488139857…

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

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

45) a_n=\dfrac{1}{n^6}, error \displaystyle <10^{−6}, \sum_{n=1}^∞\frac{1}{n^6}=\frac{π^6}{945}=1.01734306...,

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

Answer
\ln(2)

47) Find the limit as n→∞ of \dfrac{1}{n}+\dfrac{1}{n+1}+⋯+\dfrac{1}{3n}

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

48) 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 H_k=(1+\frac{1}{2}+\frac{1}{3}+⋯+\frac{1}{k}). Recall that T_k=H_k−\ln k is decreasing. Compute \displaystyle T=\lim_{k→∞}T_k to four decimal places.

\quad\Big(Hint: \displaystyle \frac{1}{k+1}<∫^{k+1}_k\frac{1}{x}\,dx.\Big)

Answer
T=0.5772...

49) [T] Complete sampling with replacement, sometimes called the coupon collector’s problem, is phrased as follows: Suppose you have 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 E(N) that it takes to draw each unique item at least once. It turns out that E(N)=N.H_N=N\left(1+\frac{1}{2}+\frac{1}{3}+⋯+\frac{1}{N}\right). Find E(N) for N=10,20, and 50.

50) [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 n cards, then the probability that the insertion will be below the card initially at the bottom (call this card B) is 1/n. Thus the expected number of top random insertions before B is no longer at the bottom is n. Once one card is below B, there are two places below B and the probability that a randomly inserted card will fall below B is 2/n. The expected number of top random insertions before this happens is n/2. The two cards below 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
The expected number of random insertions to get B to the top is n+n/2+n/3+⋯+n/(n−1). Then one more insertion puts B back in at random. Thus, the expected number of shuffles to randomize the deck is n(1+1/2+⋯+1/n).

51) Suppose a scooter can travel 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 100H_N km, where H_N=1+1/2+⋯+1/N.

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

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

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

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

Answer
The series converges for p>1 by integral test using change of variable.

55) [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 100.

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

Answer
N=e^{e^{100}}≈e^{10^{43}} terms are needed.

 


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