9.3E: Exercises for Divergence and Integral Tests

Divergence Test Problems

Consider the sequence for each series in exercises 1 - 14, if the divergence test applies, either state that \(\displaystyle \lim_{n→∞}a_n\) does not exist or find \(\displaystyle \lim_{n→∞}a_n\). If the divergence test does not apply, state why.

1) \(\displaystyle \sum_{n=1}^∞ \dfrac{n}{n+2}\)

2) \(\displaystyle \sum_{n=1}^∞\dfrac{n}{5n^2−3}\)

Answer:

\(\displaystyle \lim_{n→∞}a_n=0\). The Divergence Test does not apply.

3) \(\displaystyle \sum_{n=1}^∞\dfrac{n}{\sqrt{3n^2+2n+1}}\)

4) \(\displaystyle \sum_{n=1}^∞\dfrac{(2n+1)(n−1)}{(n+1)^2}\)

Answer:

\(\displaystyle \lim_{n→∞}a_n=2\). So the series diverges by the \(n^{\text{th}}\)-Term Test for Divergence.

5) \(\displaystyle \sum_{n=1}^∞\dfrac{(2n+1)^{2n}}{(3n^2+1)^n}\)

6) \(\displaystyle \sum_{n=1}^∞\dfrac{2^n}{3^{n/2}}\)

Answer:

\(\displaystyle \lim_{n→∞}a_n=∞\) (does not exist). So the series diverges by the \(n^{\text{th}}\)-Term Test for Divergence.

7) \(\displaystyle \sum_{n=1}^∞\dfrac{2^n+3^n}{10^{n/2}}\)

8) \(\displaystyle \sum_{n=1}^∞e^{−2/n}\)

Answer:

\(\displaystyle \lim_{n→∞}a_n=1.\) So the series diverges by the \(n^{\text{th}}\)-Term Test for Divergence.

9) \(\displaystyle \sum_{n=1}^∞\cos n\)

10) \(\displaystyle \sum_{n=1}^∞\tan n\)

Answer:

\(\displaystyle \lim_{n→∞}a_n\) does not exist. So the series diverges by the \(n^{\text{th}}\)-Term Test for Divergence.

11) \(\displaystyle \sum_{n=1}^∞\dfrac{1−\cos^2(1/n)}{\sin^2(2/n)}\)

12) \(\displaystyle \sum_{n=1}^∞\left(1−\dfrac{1}{n}\right)^{2n}\)

Answer:

\(\displaystyle \lim_{n→∞}a_n=1/e^2.\) So the series diverges by the \(n^{\text{th}}\)-Term Test for Divergence.

13) \(\displaystyle \sum_{n=1}^∞\dfrac{\ln n}{n}\)

14) \(\displaystyle \sum_{n=1}^∞\dfrac{(\ln n)^2}{\sqrt{n}}\)

Answer:

\(\displaystyle \lim_{n→∞}a_n=0.\) The Divergence Test does not apply.

\(p\)-Series Problems & Integral Test Problems

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

15) \(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt{n}}\)

16) \(\displaystyle \sum_{n=1}^∞\frac{1}{n\sqrt{n}}\)

Answer:

The series converges, since \( p=3/2>1\).

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

18) \(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt[3]{n^4}}\)

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

Answer:

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

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

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 \( \dfrac{1}{n}+\dfrac{1}{n+1}+⋯+\dfrac{1}{2n}\). (Hint: Compare to \(\displaystyle ∫^{2n}_n\frac{1}{t}\,dt.\))

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

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 \( E(N)=N.H_N=N(1+\frac{1}{2}+\frac{1}{3}+⋯+\frac{1}{N})\). 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.

Contributors

• Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.