q1s - our series converges. Integral Test : Z 1 1 x ( x +...

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MATH 153 Quiz 1 Name: 1. [4 points] Find a formula for a n in the following sequence: { a n } = ± - 2 3 , 4 9 , - 6 27 , 8 81 , - 10 243 ,... ² a n = ( - 1) n 2 n 3 n 2. [6 points] Use the Integral Test to determine whether X n =1 1 n 3 converges or diverges. Z 1 1 x 3 dx = lim b →∞ Z b 1 x - 3 dx = lim b →∞ - 1 2 x 2 ³ ³ ³ ³ b 1 = lim b →∞ - 1 2 b 2 - - 1 2 = 1 2 . Since the integral converges (is finite), we know the series must converge as well by the Integral Test.
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3. [10 points] Consider the series X n =1 1 n ( n + 1) . a. [5] Use any test to show that the series converges. I’ll show multiple ways, though you only had to show one. Comparison Test : Since 1 n ( n +1) < 1 n 2 for all n , and since n =1 1 n 2 converges (it’s a p -series with p > 1), we know by Comparison Test that our series converges. Limit Comparison Test : lim n →∞ 1 n ( n +1) 1 n 2 = lim n →∞ n 2 n 2 + n = 1 , So since n =1 1 n 2 converges, we know by Limit Comparison Test that
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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 x-1 x + 1 dx = lim b ln | x | -ln | x + 1 | b 1 = lim b ln b-ln 1-ln( b + 1) + ln 2 = ln 2 + lim b ln b b + 1 = ln 2-ln 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 n-1 n + 1 . Taking partial sums gives s n = 1 1-1 2 + 1 2-1 3 + 1 3-1 4 + ... + 1 n-1 n + 1 = 1 1-1 n + 1 . Thus, X n =1 1 n ( n + 1) = lim n s n = lim n 1-1 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.

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q1s - our series converges. Integral Test : Z 1 1 x ( x +...

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