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Unformatted text preview: APMA 111 Quiz 9 April 15, 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: For each alternating series, show that it 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. 1. [5 pts] ∞ X n =4 ( 1) n n 2 + 1 5 n Solution: This series is a perfect candidate for the Ratio Test:  a n  = n 2 + 1 5 n ⇒ a n +1 a n = ( n + 1) 2 + 1 5 n +1 · 5 n n 2 + 1 = 5 n 5 · 5 n · ( n + 1) 2 + 1 n 2 + 1 = n 2 + 2 n + 2 5( n 2 + 1) ⇒ lim n →∞ a n +1 a n = lim n →∞ n 2 + 2 n + 2 5( n 2 + 1) = lim n →∞ n 2 (1 + 2 /n + 2 /n 2 ) 5 n 2 (1 + 1 /n 2 ) = 1 + 0 + 0 5(1 + 0) = 1 5 < 1 . Since lim n →∞  a n +1 /a n  < 1, the Ratio Test implies that the series converges absolutely . 2. [5 pts] ∞ X n =2 ( 1) n n (ln n ) n Solution: This series looks like a good candidate for the Root Test:...
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This note was uploaded on 09/21/2009 for the course MATH 1234 taught by Professor Egcdd during the Spring '09 term at Aarhus Universitet.
 Spring '09
 EGCDD

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