9.3E: Exercises for Divergence and Integral Tests
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- Jul 2, 2021
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( \newcommand{\kernel}{\mathrm{null}\,}\)
Divergence Test Problems
Consider the sequence for each series in exercises 1 - 14, if the divergence test applies, either state that limn→∞an does not exist or find limn→∞an. If the divergence test does not apply, state why.
1) ∞∑n=1nn+2
2) ∞∑n=1n5n2−3
- Answer
- limn→∞an=0. The Divergence Test does not apply.
3) ∞∑n=1n√3n2+2n+1
4) ∞∑n=1(2n+1)(n−1)(n+1)2
- Answer
- limn→∞an=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
- limn→∞an=∞ (does not exist). So the series diverges by the nth-Term Test for Divergence.
7) ∞∑n=12n+3n10n/2
8) ∞∑n=1e−2/n
- Answer
- limn→∞an=1. So the series diverges by the nth-Term Test for Divergence.
9) ∞∑n=1cosn
10) ∞∑n=1tann
- Answer
- limn→∞an does not exist. So the series diverges by the nth-Term Test for Divergence.
11) ∞∑n=11−cos2(1/n)sin2(2/n)
12) ∞∑n=1(1−1n)2n
- Answer
- limn→∞an=1/e2. So the series diverges by the nth-Term Test for Divergence.
13) ∞∑n=1lnnn
14) ∞∑n=1(lnn)2√n
- Answer
- limn→∞an=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=11√n
16) ∞∑n=11n√n
- Answer
- The series converges, since p=3/2>1.
17) ∞∑n=113√n2
18) ∞∑n=113√n4
- Answer
- The series converges, since p=4/3>1.
19) ∞∑n=1nenπ
20) ∞∑n=1nπn2e
- 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) ∞∑n=11√n+5
22) ∞∑n=113√n+5
- Answer
- The series diverges by the Integral Test since ∫∞1dx(x+5)1/3 can be shown to diverge.
23) ∞∑n=21nlnn
24) ∞∑n=1n1+n2
- Answer
- The series diverges by the Integral Test since ∫∞1x1+x2dx can be shown to diverge.
25) ∞∑n=1en1+e2n
26) ∞∑n=12n1+n4
- Answer
- The series converges by the Integral Test since ∫∞12x1+x4dx converges.
27) ∞∑n=21nln2n
Express the sums in exercises 28 - 31 as p-series and determine whether each converges.
28) ∞∑n=12−lnn (Hint: 2−lnn=1nln2.)
- Answer
- 2−lnn=1/nln2. Since p=ln2<1, this series diverges by the p-series test.
29) ∞∑n=13−lnn (Hint: 3−lnn=1nln3.)
30) n∑n=12−2lnn
- Answer
- 2−2lnn=1/n2ln2. Since p=2ln2−1<1, this series diverges by the p-series test.
31) ∞∑n=1n3−2lnn
In exercises 32 - 35, use the estimate RN≤∫∞Nf(t)dt to find a bound for the remainder RN=∞∑n=1an−N∑n=1an where an=f(n).
32) 1000∑n=11n2
- Answer
- R1000≤∫∞1000dtt2=limb→∞−1t|b1000=limb→∞(−1b+11000)=0.001
33) 1000∑n=11n3
34) 1000∑n=111+n2
- Answer
- R1000≤∫∞1000dt1+t2=limb→∞(tan−1b−tan−1(1000))=π/2−tan−1(1000)≈0.000999
35) 100∑n=1n2n
[T] In exercises 36 - 40, find the minimum value of N such that the remainder estimate ∫∞N+1f(x)dx<RN<∫∞Nf(x)dx guarantees that N∑n=1an estimates ∞∑n=1an, accurate to within the given error.
36) an=1n2, error <10−4
- Answer
- RN<∫∞Ndxx2=1/N,forN>104
37) an=1n1.1, error <10−4
38) an=1n1.01, error <10−4
- Answer
- RN<∫∞Ndxx1.01=100N−0.01,forN>10600
39) an=1nln2n, error <10−3
40) an=11+n2, error <10−3
- Answer
- RN<∫∞Ndx1+x2=π/2−tan−1(N),forN>tan(π/2−10−3)≈1000
In exercises 41 - 45, find a value of N such that RN is smaller than the desired error. Compute the corresponding sum N∑n=1an and compare it to the given estimate of the infinite series.
41) an=1n11, error <10−4,∞∑n=11n11=1.000494…
42) an=1en, error <10−5,∞∑n=11en=1e−1=0.581976…
- Answer
- RN<∫∞Ndxex=e−N,forN>5ln(10), okay if N=12;12∑n=1e−n=0.581973.... Estimate agrees with 1/(e−1) to five decimal places.
43) an=1en2, error <10−5,∞∑n=11en2=0.40488139857…
44) an=1n4, error <10−4,∞∑n=11n4=π490=1.08232...
- Answer
- RN<∫∞Ndxx4=4/N3,forN>(4.104)1/3, okay if N=35; 35∑n=11n4=1.08231…. Estimate agrees with the sum to four decimal places.
45) an=1n6, error <10−6,∞∑n=11n6=π6945=1.01734306...,
46) Find the limit as n→∞ of 1n+1n+1+⋯+12n. (Hint: Compare to ∫2nn1tdt.)
- Answer
- ln(2)
47) Find the limit as n→∞ of 1n+1n+1+⋯+13n
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 Hk=(1+12+13+⋯+1k). Recall that Tk=Hk−lnk is decreasing. Compute T=limk→∞Tk to four decimal places.
(Hint: 1k+1<∫k+1k1xdx.)
- 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.HN=N(1+12+13+⋯+1N). 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 100HN km, where HN=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 bn=an+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 ∞∑n=1nen within an error smaller than 0.0001.
54) Does ∞∑n=21n(lnn)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 N∑n=11n. 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 N∑n=21nlnn. 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=ee100≈e1043 terms are needed.