Unformatted text preview: p = 1, so assume that p ± = 1. Z ∞ 1 1 x p dx = lim D →∞ x 1p 1p ﬂ ﬂ ﬂ ﬂ D 1 = lim D →∞ D 1p 1p1 1p . If p > 1 then 1p < 0 and lim D →∞ D 1p = 0, so the integral converges. If 0 < p < 1 then 1p > 0 and lim D →∞ D 1p = ∞ , 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 pseries 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 pseries, ∑ ∞ n =1 5 /n 4 converges also....
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 Spring '07
 JonathanRogawski
 Math, Calculus, lim

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