121616949-math.261 - p = 1 so assume that p ± = 1 Z ∞ 1...

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10.3 The Integral Test 247 which we can rewrite as Z 1 1 x dx = . So these two examples taken together indicate that we can prove that a series converges or prove that it diverges with a single calculation of an improper integral. This is known as the integral test , which we state as a theorem. THEOREM 10.23 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 X n =1 a n converges if and only if the improper integral Z 1 f ( 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, lim n →∞ 1 /n p = 0, so the series diverges. For positive values of p we can determine precisely which series converge. THEOREM 10.24 A p -series with p > 0 converges if and only if p > 1. Proof. We use the integral test; we have already done
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Unformatted text preview: p = 1, so assume that p ± = 1. Z ∞ 1 1 x p dx = lim D →∞ x 1-p 1-p fl fl fl fl D 1 = lim D →∞ D 1-p 1-p-1 1-p . If p > 1 then 1-p < 0 and lim D →∞ D 1-p = 0, so the integral converges. If 0 < p < 1 then 1-p > 0 and lim D →∞ D 1-p = ∞ , so the integral diverges. EXAMPLE 10.25 Show that ∞ X n =1 1 n 3 converges. We could of course use the integral test, but now that we have the theorem we may simply note that this is a p-series with p > 1. EXAMPLE 10.26 Show that ∞ X n =1 5 n 4 converges. We know that if ∑ ∞ n =1 1 /n 4 converges then ∑ ∞ n =1 5 /n 4 also converges, by theorem 10.17 . Since ∑ ∞ n =1 1 /n 4 is a convergent p-series, ∑ ∞ n =1 5 /n 4 converges also....
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