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I'm trying to prove whether the following series is convergent, divergent, or that there is not enough info.

If the series ∑ an is convergent and has positive terms, what is the series below?

$$ \sum{(-1)^na_n}$$

I know that the series a_n is absolutely convergent, so the alternating one should be convergent, but I"m not sure how to prove it. The limit of an as n goes to infinity is 0, so an should be decreasing eventually, so it should be convergent by the Alternating Series Test.

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  • $\begingroup$ Leibnitz theorem proves that $\endgroup$ Commented May 20, 2014 at 14:23
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    $\begingroup$ Use $\Bigl|\sum_{n=k}^{k+m} (-1)^n a_n\Bigr|\le \sum_{n=k}^{k+m}|a_n|$. $\endgroup$ Commented May 20, 2014 at 14:26
  • $\begingroup$ Thanks, David, that works! $\endgroup$ Commented May 20, 2014 at 14:28

2 Answers 2

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The general rule is that if $\sum|x_n|$ converges, then $\sum x_n$ converges. This is a simple application of that fact.

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By the convergence of $\sum_0^\infty a_k$, the sequence $(s_n)$, where $s_n=\sum_0^n (-1)^k a_k$ is a Cauchy sequence.

Proof: If $m\lt n$, then $|s_n-s_m|\le a_{m+1}+\cdots +a_n$.

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  • $\begingroup$ I haven't studied Cauchy sequences, is there another way to do this? $\endgroup$ Commented May 20, 2014 at 14:27
  • $\begingroup$ @ElectronicGeek If you know about absolute convergence, you should have studied Cauchy sequences, since usually they're what you use to prove that absolute convergence implies convergence. $\endgroup$ Commented May 20, 2014 at 14:30
  • $\begingroup$ One can get it from Bolzano-Weierstrass. $\endgroup$ Commented May 20, 2014 at 14:30

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