# Convergence of infinite series $\sum_{k=2}^{\infty}\frac{(-1)^k}{k\ln(k)}$

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!

• alternating series test – user10354138 Oct 16 '18 at 8:27
• Leibniz criterion – Peter Szilas Oct 16 '18 at 8:28

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.

• Thank you! This is really an interesting way to do it. – dogthepeter Oct 24 '18 at 2:34

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.

• Many thanks for pointing it out! I wasn't aware of the criterion at first place. It is surely a useful and interesting result! – dogthepeter Oct 24 '18 at 2:36

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.

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