Sequences and Series

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268 Chapter 11 Sequences and SeriesTherectanglesthistimeareabovethecurve, thatis,eachrectanglecompletelycontainsthe corresponding area under the curve. This means thats n = 1 1 + 1 2 + 1 3 +···+ 1 n > ∫ n+111∣ ∣∣x dx = lnx n+1= ln(n+1).1As n gets bigger, ln(n+1) goes to infinity, so the sequence of partial sums s n must alsogo to infinity, so the harmonic series diverges.The important fact that clinches this example is thatwhich we can rewrite as∫ n+11lim dx = ∞,n→∞1 x∫ ∞11dx = ∞.xSo these two examples taken together indicate that we can prove that a series convergesor prove that it diverges with a single calculation of an improper integral. This is knownas the integral test, which we state as a theorem.THEOREM 11.23 Suppose that f(x) > 0 and is decreasing on the infinite interval∞∑[k,∞) (for some k ≥ 1) and that a n = f(n). Then the series a n converges if and onlyif the improper integral∫ ∞1f(x)dx converges.The two examples we have seen are called p-series; a p-series is any series of the form∑1/n p . If p ≤ 0, limn→∞ 1/np ≠ 0, so the series diverges. For positive values of p we candetermine precisely which series converge.THEOREM 11.24 A p-series with p > 0 converges if and only if p > 1.Proof. We use the integral test; we have already done p = 1, so assume that p ≠ 1.n=1∫ ∞11dx = limxp D→∞x 1−p1−p∣D1D 1−p= limD→∞ 1−p − 11−p .If p > 1 then 1−p < 0 and limD→∞ D1−p = 0, so the integral converges. If 0 0 and limD→∞ D1−p = ∞, so the integral diverges.
EXAMPLE 11.25 Show that∞∑n=11n 3 converges.11.3 The Integral Test 269We could of course use the integral test, but now that we have the theorem we may simplynote that this is a p-series with p > 1.EXAMPLE 11.26 Show thatWe know that if∞∑n=1∞∑1/n 4 converges thenn=1∞∑1/n 4 is a convergent p-series,n=1EXAMPLE 11.27 Show that5n 4 converges.∞∑5/n 4 also converges, by theorem 11.17. Sincen=1∞∑5/n 4 converges also.n=1∞∑n=15√ ndiverges.This also follows from theorem 11.17: Sincediverges, and so does∞∑n=15√ n.∞∑n=11√ nis a p-series with p = 1/2 < 1, itSince it is typically difficult to compute the value of a series exactly, a good approximationis frequently required. In a real sense, a good approximation is only as good aswe know it is, that is, while an approximation may in fact be good, it is only valuable inpractice if we can guarantee its accuracy to some degree. This guarantee is usually easyto come by for series with decreasing positive terms.EXAMPLE 11.28 Approximate ∑ 1/n 2 to two decimal places.N∑Referring to figure 11.2, if we approximate the sum by 1/n 2 , the error we make isthe total area of the remaining rectangles, all of which lie under the curve 1/x 2 from x = Nout to infinity. So we know the true value of the series is larger than the approximation,and no bigger than the approximation plus the area under the curve from N to infinity.Roughly, then, we need to find N so that∫ ∞N1dx < 1/100.x2 n=1
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