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Unformatted text preview: Solutions to Midterm 1 Test the following series for convergence: 1. X n =0 ( 2 3 ) n . Solution: This is a geometric series with the ratio r = 2 3 &lt; 1, so it converges. Grading: 0 points if the answer is divergent; 4 points if no check  r  &lt; 1 is done. 2. X n =1 n 2 1 n n 4 . Solution: a) We can use the Comparison Test: &lt; n 2 1 n n 4 &lt; n 2 n 4 = 1 n 2 . The series n =1 1 n 2 is a pseries with p = 2 &gt; 1, so it converges, therefore the initial series converges by the Comparison Test. Grading: 2 points for the convergence of 1 n 2 , 6 points for the limit computation in limit comparison test, 2 points for the conclusion. b) X n =1 n 2 1 n n 4 = X n =1 1 n 2 X n =1 1 n 5 . Since both series are converging pseries with p = 2 , 5 &gt; 1, the initial series is convergent as well. Grading: 4 points for the presentation as the difference of two series, 4 points for the pseries test, 2 points for the conclusion 3. X n =1 n 3 + n n 5 1 . 1 Solution: We can use the Limit Comparison Test. Let a n = n 3 + n n 5 1 ,b n = 1 n 2 , then lim n a n b n = lim n n 2 ( n 3 + n ) n 5 1 = lim n 1 + 1 /n 2 1 1 /n 5 = 1 , and a n ,b n &gt; 0. The series n =2 b n = n =2 1 n 2 is a pseries with p = 2 &gt; 1, so it converges, therefore the initial series converges by the Limit Comparison...
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 Fall '07
 GuanYuShi
 Calculus, Geometric Series, Mathematical Series, lim, UDV

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