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In a recent assignment, as an intermediate step, I need to show that $\sum_{k=2}^{\infty}\frac{(-1)^k}{k\ln(k)}$ converges. It is not hard to see that $\sum_{k=2}^{\infty}\frac{1}{k\ln(k)}$ is divergent, therefore I think one has to deal with the sum of alternating sequence directly. However, I am stuck and don't know how to proceed. Thanks in advance for anyone that is kind to help!

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    $\begingroup$ alternating series test $\endgroup$ – user10354138 Oct 16 '18 at 8:27
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    $\begingroup$ Leibniz criterion $\endgroup$ – Peter Szilas Oct 16 '18 at 8:28
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If, for some reason, you do not want to use the Leibniz criterion, you can group your sum in pairs: $$ a_n = \frac{(-1)^{2n}}{2n\log(2n)}+\frac{(-1)^{2n+1}}{(2n+1)\log(2n+1)} $$ is positive, and $a_n < \frac{1}{n^2}$ so $\sum a_n$ converges.

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  • $\begingroup$ Thank you! This is really an interesting way to do it. $\endgroup$ – dogthepeter Oct 24 '18 at 2:34
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The series converges, and it can be proven using the Leibniz criterion for alternating series.

The criterion analyzes sums of the form $$\sum_{n=1}^\infty (-1)^n a_n$$ where $a_i\geq 0$. The criterion says that if $\lim_{n\to\infty}a_n = 0$ and the sequence $\{a_n\}$ is decreasing, then the sum converges. In your case, $a_k=\frac{1}{k\ln k}$ which satisfies both conditions (it's decreasing and has a limit of $0$), so the series converges.

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  • $\begingroup$ Many thanks for pointing it out! I wasn't aware of the criterion at first place. It is surely a useful and interesting result! $\endgroup$ – dogthepeter Oct 24 '18 at 2:36
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You may also couple adjacent terms: $$ 0<\frac{1}{2k\log(2k)}-\frac{1}{(2k+1)\log(2k+1)} = \frac{2k\log\left(1+\frac{1}{2k}\right)+\log(2k+1)}{2k(2k+1)\log(2k)\log(2k+1)}<\frac{1}{k^2} $$ and recall that $\sum_{k\geq 1}\frac{1}{k^2}$ is convergent.

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  • $\begingroup$ Thank you! I think the answer is similar to the one provided by GEdger. Truly interesting! $\endgroup$ – dogthepeter Oct 24 '18 at 2:35

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