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ConditionalConvergence

# ConditionalConvergence - S = ∞ X n =1 ± 1 2(2 n-1-1 4 n...

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CONDITIONAL CONVERGENCE Here is a concrete example of conditional convergence. Recall that the series S = X n =1 ( - 1) n +1 n converges (by the alternating series test), but not absolutely. (The fact that it converges to a known limit ln 2 is irrelevent; it is only important to know that S 6 = 0 , for the following argument to make sense.) Writing the terms of S out for clarity, we have S = 1 - 1 2 + 1 3 - 1 4 + 1 5 - 1 6 + 1 7 - 1 8 + 1 9 - 1 10 + ··· We now permute the terms in the sequence as follows. After the ﬁrst term 1 , we shift all even fractions back so they come in groups of two, with the odd fractions interspersed between them, as follows. S 0 = 1 - 1 2 - 1 4 + 1 3 - 1 6 - 1 8 + 1 5 - 1 10 - 1 12 + 1 7 - 1 14 - 1 16 + ··· We break this rearrangement into blocks of three terms, S 0 = ± 1 - 1 2 - 1 4 ² + ± 1 3 - 1 6 - 1 8 ² + ± 1 5 - 1 10 - 1 12 ² + ± 1 7 - 1 14 - 1 16 ² + ··· In summation notation, what we have shown is that S 0 = X n =1 ± 1 2 n - 1 - 1 2(2 n - 1) - 1 4 n ² is a rearrangement of S = n =1 ( - 1) n +1 n . But the ﬁrst two terms in each block of three simpliﬁes: 1 2 n - 1 - 1 2(2 n - 1) = 1 2(2 n - 1) , and so we can evaluate
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Unformatted text preview: S = ∞ X n =1 ± 1 2(2 n-1)-1 4 n ² . Expressing the difference of fractions over a common denominator, we have the terms of the series S are 4 n-2(2 n-1) 8 n (2 n-1) = 1 4 n (2 n-1) . By the comparison test, we have, since 2 n-1 ≥ 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 n-1-1 2 n ² . But, writing out the terms in the summation, we see that ∞ X n =1 ± 1 2 n-1-1 2 n ² = ± 1-1 2 ² + ± 1 3-1 4 ² + ± 1 5-1 6 ² + ··· = S. Hence, S = 1 2 S , although S is a rearrangement of S . Finally, note that that 1 n (2 n-1) ≥ 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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