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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 b b + 1 = ln 2ln 1 = ln 2 . Since the integral converges, so does the series. b. [5] Rewrite the series as a telescoping series and nd its limit. X n =1 1 n ( n + 1) = X n =1 1 n1 n + 1 . Taking partial sums gives s n = 1 11 2 + 1 21 3 + 1 31 4 + ... + 1 n1 n + 1 = 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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This note was uploaded on 07/26/2011 for the course MATH 153 taught by Professor Rempe during the Spring '08 term at Ohio State.
 Spring '08
 REMPE
 Math

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