Unformatted text preview: S = ∞ X n =1 ± 1 2(2 n1)1 4 n ² . Expressing the difference of fractions over a common denominator, we have the terms of the series S are 4 n2(2 n1) 8 n (2 n1) = 1 4 n (2 n1) . By the comparison test, we have, since 2 n1 ≥ n , that this is ≤ 1 4 n 2 , and hence the series S converges. We may therefore use the limit theorems to pull out a factor of 1 2 : S = 1 2 ∞ X n =1 ± 1 2 n11 2 n ² . But, writing out the terms in the summation, we see that ∞ X n =1 ± 1 2 n11 2 n ² = ± 11 2 ² + ± 1 31 4 ² + ± 1 51 6 ² + ··· = S. Hence, S = 1 2 S , although S is a rearrangement of S . Finally, note that that 1 n (2 n1) ≥ 1 2 n 2 , which means that S ≥ 1 8 ∑ 1 n 2 > . Hence, S = 1 2 S is not equal to S ....
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 Fall '10
 Prof.KatrinWehrheim
 n=1

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