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Unformatted text preview: Test 4 Solutions 1 . Use the integral test to determine if n =1 1 n 3 converges. Compute 1 1 x 3 dx = lim t t 1 1 x 3 dx = lim t  1 2 x 2 t 1 = lim t 1 2 1 2 t 2 = 1 2 . There fore, the series converges. 2 . Use the comparison test to determine if n =1 1 n 2 + 3 converges. Note 1 n 2 + 3 < 1 n 2 for n 1. Therefore, since n =1 1 n 2 converges (it is a pseries with p = 2), the series n =1 1 n 2 + 3 converges as well. 3 . Use the limit comparison test to determine if n =1 1 n 2 + 1 converges. Since 1 n 2 + 1 1 n 2 = 1 n , we compare with 1 n . So we compute lim n 1 n 2 +1 1 n = lim n 1 n 2 + 1 n 1 = lim n n n 2 + 1 = lim n n n 2 + 1 = 1 = 1 . Now the series 1 n diverges (it is the harmonic series, a pseries with p = 1). So by the limitcomparison test, so does n =1 1 n 2 + 1 . (The limit needs to be a positive, finite number for the limitcomparison test to work.) 4 . Determine if the series n =1 ( 1) n +1 n converges. State which test you used....
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 Fall '08
 VALDIMARSSON

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