227_series_solns[1]

227_series_solns[1] - Math 227.04 Solutions for Written...

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Unformatted text preview: Math 227.04 Solutions for Written 4/22 hw 7.6.8 First, we need to check that P 1 n =1 ( & 1) n ln n n & ln n meets the conditions of the Alternating Series Test. Setting u n = ln n n & ln n , we must show ( i ) u n > and ( ii ) u n decreases with n . Let f ( x ) = ln x n & ln x : Clearly f ( x ) > for x 1 : Also, f ( x ) = 1 x ( x & ln x ) & (ln x ) & 1 & 1 x ( x & ln x ) 2 = 1 & ln x ( x & ln x ) 2 < for x > e: Since 3 > e; f ( x ) is a decreasing function of x on the interval (3 ; 1 ) This means the u n = f ( n ) decreases as for n 3 , so we can apply the AST. Using L&Hospital&s rule, lim n !1 u n = lim x !1 ln x x & ln x = lim x !1 1 x 1 & 1 x = 0 : Thus the series converges. Since ln n n & ln n > 1 n when n 3 ; we see that P 1 n =1 ( & 1) n ln n n & ln n diverges. Conclusion: our original series is CONDITIONALLY COVERGENT. 7.6.50 We can apply Alternating Series Remainder Theorem (Theorem 15 of Section 7.6) to guarantee that the magnitude of the error in the approximation...
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227_series_solns[1] - Math 227.04 Solutions for Written...

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