Although convergence of ∫N∞f(x)dx∫N∞f(x)dx implies convergence of the related series ∑n=1∞an,∑n=1∞an, it does not imply that the value of the integral and the series are the same. They may be different, and often are. For example,

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For each of the following series, if the divergence test applies, either state that limn→∞anlimn→∞an does not exist or find limn→∞an.limn→∞an. If the divergence test does not apply, state why.

In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums {Sk}{Sk} and showing that S2k>1+k/2S2k>1+k/2 for all positive integers k.k. In this section we use a different technique to prove the divergence of the harmonic series. This technique is important because it is used to prove the divergence or convergence of many other series. This test, called the integral test, compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.

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[T] Complete sampling with replacement, sometimes called the coupon collector’s problem, is phrased as follows: Suppose you have NN 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)E(N) that it takes to draw each unique item at least once. It turns out that E(N)=NE(N)=N. HN=N(1+12+13+⋯+1N)HN=N(1+12+13+⋯+1N). Find E(N)E(N) for N=10,20,and50N=10,20,and50.

a n = 1 − cos 2 ( 1 / n ) sin 2 ( 2 / n ) a n = 1 − cos 2 ( 1 / n ) sin 2 ( 2 / n )

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

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For each of the following series, apply the divergence test. If the divergence test proves that the series diverges, state so. Otherwise, indicate that the divergence test is inconclusive.

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It is important to note that the converse of this theorem is not true. That is, if limn→∞an=0,limn→∞an=0, we cannot make any conclusion about the convergence of ∑n=1∞an.∑n=1∞an. For example, limn→∞(1/n)=0,limn→∞(1/n)=0, but the harmonic series ∑n=1∞1/n∑n=1∞1/n diverges. In this section and the remaining sections of this chapter, we show many more examples of such series. Consequently, although we can use the divergence test to show that a series diverges, we cannot use it to prove that a series converges. Specifically, if an→0,an→0, the divergence test is inconclusive.

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We illustrate Remainder Estimate from the Integral Test in Figure 5.15. In particular, by representing the remainder RN=aN+1+aN+2+aN+3+⋯RN=aN+1+aN+2+aN+3+⋯ as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by ∫N∞f(x)dx∫N∞f(x)dx and bounded below by ∫N+1∞f(x)dx.∫N+1∞f(x)dx. In other words,

Since {Sk}{Sk} is increasing and bounded, by the Monotone Convergence Theorem, it converges. Therefore, the series ∑n=1∞1/n2∑n=1∞1/n2 converges.

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).Hk=(1+12+13+⋯+1k). Recall that Tk=Hk−lnkTk=Hk−lnk is decreasing. Compute T=limk→∞TkT=limk→∞Tk to four decimal places. (Hint: 1k+1<∫kk+11xdx1k+1<∫kk+11xdx.)

Use the estimate RN≤∫N∞f(t)dtRN≤∫N∞f(t)dt to find a bound for the remainder RN=∑n=1∞an−∑n=1NanRN=∑n=1∞an−∑n=1Nan where an=f(n).an=f(n).

Suppose ∑n=1∞an∑n=1∞an is a convergent series with positive terms. Suppose there exists a function ff satisfying the following three conditions:

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[T] Use the remainder estimate and integration by parts to approximate ∑n=1∞n/en∑n=1∞n/en within an error smaller than 0.0001.0.0001.

In the previous section, we determined the convergence or divergence of several series by explicitly calculating the limit of the sequence of partial sums {Sk}.{Sk}. In practice, explicitly calculating this limit can be difficult or impossible. Luckily, several tests exist that allow us to determine convergence or divergence for many types of series. In this section, we discuss two of these tests: the divergence test and the integral test. We will examine several other tests in the rest of this chapter and then summarize how and when to use them.

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Now consider the series ∑n=1∞1/n2.∑n=1∞1/n2. We show how an integral can be used to prove that this series converges. In Figure 5.13, we sketch a sequence of rectangles with areas 1,1/22,1/32,…1,1/22,1/32,… along with the function f(x)=1/x2.f(x)=1/x2. From the graph we see that

We can extend this idea to prove convergence or divergence for many different series. Suppose ∑n=1∞an∑n=1∞an is a series with positive terms anan such that there exists a continuous, positive, decreasing function ff where f(n)=anf(n)=an for all positive integers. Then, as in Figure 5.14(a), for any integer k,k, the kthkth partial sum SkSk satisfies

[T] Find the minimum value of NN such that the remainder estimate ∫N+1∞f<∫N∞f∫N+1∞f<∫N∞f guarantees that ∑n=1Nan∑n=1Nan estimates ∑n=1∞an,∑n=1∞an, accurate to within the given error.

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Suppose a scooter can travel 100100 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 100HN100HN km, where HN=1+1/2+⋯+1/N.HN=1+1/2+⋯+1/N.

a n = ( 2 n + 1 ) 2 n ( 3 n 2 + 1 ) n a n = ( 2 n + 1 ) 2 n ( 3 n 2 + 1 ) n

Since limk→∞ln(k+1)=∞,limk→∞ln(k+1)=∞, we see that the sequence of partial sums {Sk}{Sk} is unbounded. Therefore, {Sk}{Sk} diverges, and, consequently, the series ∑n=1∞1n∑n=1∞1n also diverges.

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If p>0,p>0, then f(x)=1/xpf(x)=1/xp is a positive, continuous, decreasing function. Therefore, for p>0,p>0, we use the integral test, comparing

DivergenceCalculus

Therefore, if ∫1∞f(x)dx∫1∞f(x)dx converges, then the sequence of partial sums {Sk}{Sk} is bounded. Since {Sk}{Sk} is an increasing sequence, if it is also a bounded sequence, then by the Monotone Convergence Theorem, it converges. We conclude that if ∫1∞f(x)dx∫1∞f(x)dx converges, then the series ∑n=1∞an∑n=1∞an also converges. On the other hand, from Figure 5.14(b), for any integer k,k, the kthkth partial sum SkSk satisfies

a n = ( 2 n + 1 ) ( n − 1 ) ( n + 1 ) 2 a n = ( 2 n + 1 ) ( n − 1 ) ( n + 1 ) 2

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For each of the following series, use the integral test to determine whether the series converges or diverges. Assume that all conditions for the integral test are met.

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[T] A fast computer can sum one million terms per second of the divergent series ∑n=2N1nlnn.∑n=2N1nlnn. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed 100.100.

be the remainder when the sum of an infinite series is approximated by the NthNth partial sum, how large is RN?RN? For some types of series, we are able to use the ideas from the integral test to estimate RN.RN.

Therefore, if ∑n=1∞an∑n=1∞an converges, the nthnth term an→0an→0 as n→∞.n→∞. An important consequence of this fact is the following statement:

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[T] Suppose a computer can sum one million terms per second of the divergent series ∑n=1N1n.∑n=1N1n. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed 100.100.

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Suppose ∑n=1∞an∑n=1∞an is a series with positive terms an.an. Suppose there exists a function ff and a positive integer NN such that the following three conditions are satisfied:

[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 nn cards, then the probability that the insertion will be below the card initially at the bottom (call this card B)B) is 1/n.1/n. Thus the expected number of top random insertions before BB is no longer at the bottom is n. Once one card is below B,B, there are two places below BB and the probability that a randomly inserted card will fall below BB is 2/n.2/n. The expected number of top random insertions before this happens is n/2.n/2. The two cards below BB 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.

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In the following exercises, find a value of NN such that RNRN is smaller than the desired error. Compute the corresponding sum ∑n=1Nan∑n=1Nan and compare it to the given estimate of the infinite series.

We know the p-series converges if p=2p=2 and diverges if p=1.p=1. What about other values of p?p? In general, it is difficult, if not impossible, to compute the exact value of most pp-series. However, we can use the tests presented thus far to prove whether a pp-series converges or diverges.

Suppose we know that a series ∑n=1∞an∑n=1∞an converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum ∑n=1Nan∑n=1Nan where NN is any positive integer. The question we address here is, for a convergent series ∑n=1∞an,∑n=1∞an, how good is the approximation ∑n=1Nan?∑n=1Nan? More specifically, if we let

To illustrate how the integral test works, use the harmonic series as an example. In Figure 5.12, we depict the harmonic series by sketching a sequence of rectangles with areas 1,1/2,1/3,1/4,…1,1/2,1/3,1/4,… along with the function f(x)=1/x.f(x)=1/x. From the graph, we see that

If limk→∞∫1k+1f(x)dx=∞,limk→∞∫1k+1f(x)dx=∞, then {Sk}{Sk} is an unbounded sequence and therefore diverges. As a result, the series ∑n=1∞an∑n=1∞an also diverges. We conclude that if ∫1∞f(x)dx∫1∞f(x)dx diverges, then ∑n=1∞an∑n=1∞an diverges.

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