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227_series_solns[1]

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

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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 > 0 and ( ii ) u n decreases with n . Let f ( x ) = ln x n ° ln x : Clearly f ( x ) > 0 for x ± 1 : Also, f 0 ( x ) = 1 x ( x ° ln x ) ° (ln x ) ° 1 ° 1 x ± ( x ° ln x ) 2 = 1 ° ln x ( x ° ln x ) 2 < 0 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 ² ² ² ( ° 1) n ln n n ° ln n ² ² ² 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 1 X k =0 ( ° 1) k 1 k !

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

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