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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) n=1nenπ

20) n=1nπn2e

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

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

21) n=11n+5

22) n=113n+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=12lnn (Hint: 2lnn=1nln2.)

Answer
2lnn=1/nln2. Since p=ln2<1, this series diverges by the p-series test.

29) n=13lnn (Hint: 3lnn=1nln3.)

30) nn=122lnn

Answer
22lnn=1/n2ln2. Since p=2ln21<1, this series diverges by the p-series test.

31) n=1n32lnn

In exercises 32 - 35, use the estimate RNNf(t)dt to find a bound for the remainder RN=n=1anNn=1an where an=f(n).

32) 1000n=11n2

Answer
R10001000dtt2=limb1t|b1000=limb(1b+11000)=0.001

33) 1000n=11n3

34) 1000n=111+n2

Answer
R10001000dt1+t2=limb(tan1btan1(1000))=π/2tan1(1000)0.000999

35) 100n=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 Nn=1an estimates n=1an, accurate to within the given error.

36) an=1n2, error <104

Answer
RN<Ndxx2=1/N,forN>104

37) an=1n1.1, error <104

38) an=1n1.01, error <104

Answer
RN<Ndxx1.01=100N0.01,forN>10600

39) an=1nln2n, error <103

40) an=11+n2, error <103

Answer
RN<Ndx1+x2=π/2tan1(N),forN>tan(π/2103)1000

In exercises 41 - 45, find a value of N such that RN is smaller than the desired error. Compute the corresponding sum Nn=1an and compare it to the given estimate of the infinite series.

41) an=1n11, error <104,n=11n11=1.000494

42) an=1en, error <105,n=11en=1e1=0.581976

Answer
RN<Ndxex=eN,forN>5ln(10), okay if N=12;12n=1en=0.581973.... Estimate agrees with 1/(e1) to five decimal places.

43) an=1en2, error <105,n=11en2=0.40488139857

44) an=1n4, error <104,n=11n4=π490=1.08232...

Answer
RN<Ndxx4=4/N3,forN>(4.104)1/3, okay if N=35; 35n=11n4=1.08231. Estimate agrees with the sum to four decimal places.

45) an=1n6, error <106,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=Hklnk is decreasing. Compute T=limkTk 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/(n1). 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 Nn=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 Nn=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=ee100e1043 terms are needed.

 


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