111_quiz10_spr08_key

111_quiz10_spr08_key - APMA 111 NAME Quiz 10 SECTION#2(11...

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APMA 111 Quiz 10 April 22, 2008 NAME: SECTION: #2 (11 am) #6 (10 am) On my honor as a student, I have neither given nor received aid on this quiz. SIGNATURE: 1. [5 pts] Show that the following series is absolutely convergent, conditionally convergent, or divergent. Clearly identify which test you are using, and (if necessary) verify that the conditions for the test are met. X n =1 n 2 2 n - 1 ( - 5) n Solution: This series is a perfect candidate for the Ratio Test: | a n | = n 2 2 n - 1 5 n ± ± ± ± a n +1 a n ± ± ± ± = ( n + 1) 2 2 ( n +1) - 1 5 n +1 · 5 n n 2 2 n - 1 = ² n + 1 n ³ 2 · ± ± 5 n 5 · ± ± 5 n · 2 · ² ² ² 2 n - 1 ² ² ² 2 n - 1 = 2 5 ² n + 1 n ³ 2 = 2 5 (1 + 1 /n ) 2 lim n →∞ ± ± ± ± a n +1 a n ± ± ± ± = 2 5 lim n →∞ (1 + 1 /n ) 2 = 2 5 (1 + 0) 2 = 2 5 < 1 . Since lim n →∞ | a n +1 /a n | < 1, the Ratio Test implies that the series converges absolutely . 2. [5 pts] Find the radius of convergence and the interval of convergence of the power series below. X
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111_quiz10_spr08_key - APMA 111 NAME Quiz 10 SECTION#2(11...

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