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Unformatted text preview: our series converges. Integral Test : Z ∞ 1 1 x ( x + 1) dx = lim b →∞ Z b 1 1 x1 x + 1 dx = lim b →∞ ln  x  ln  x + 1  ± ± ± b 1 = lim b →∞ ln bln 1ln( b + 1) + ln 2 = ln 2 + lim b →∞ ln ² b b + 1 ³ = ln 2ln 1 = ln 2 . Since the integral converges, so does the series. b. [5] Rewrite the series as a telescoping series and ﬁnd its limit. ∞ X n =1 1 n ( n + 1) = ∞ X n =1 1 n1 n + 1 . Taking partial sums gives s n = ² 1 11 2 ³ + ² 1 21 3 ³ + ² 1 31 4 ³ + ... + ² 1 n1 n + 1 ³ = 1 11 n + 1 . Thus, ∞ X n =1 1 n ( n + 1) = lim n →∞ s n = lim n →∞ 11 n + 1 = 1 ....
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 Spring '08
 REMPE
 Math, Mathematical Series, lim, lim ln

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