Solutions_6_W

# Answer converges solution since 1 cos n 1 we have a n

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Answer: Converges. Solution: Since - 1 cos n 1, we have a n = cos 2 n 1 + n 2 < 1 1 + n 2 < 1 n 2 . The series n =1 1 n 2 converges (it is a p -series with p = 2), so by Comparison Test the series n =1 a n converges as well. 31. (10 points) Find the values for p such that the following series is convergent. X n =2 1 n (ln n ) p Answer: Converges for p > 1. Solution: Let us use the integral test. The convergence of the series is equivalent to the convergence of the similar integral: Z 2 dx x (ln x ) p . Let us make a change of variables u = ln x , then du = dx x , and Z 2 dx x (ln x ) p = Z ln 2 du u p . Since for p 6 = 1 Z T ln 2 du u p = 1 1 - p u 1 - p | T ln 2 = 1 1 - p ( T 1 - p - (ln 2) 1 - p ) and Z T ln 2 du u = (ln u ) | T ln 2 = ln T - ln ln 2 , these definite integrals have finite limit at T → ∞ (i.e. the indefinite integrals converge) if and only if 1 - p < 0 p > 1. 3
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• Fall '07
• Guan-YuShi

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