# How to prove the convergence of $\sum_{n=1}^\infty (-1)^n\frac{\ln n}{n}$?

$$\sum_{n=1}^\infty (-1)^n\frac{\ln n}{n}$$

The above series is the one I want to prove convergent. I took $\frac{\ln n}{n}$ but I didn't find anything that I could do with it. I tried to compare it with another series, didn't find a good one. Any idea for this?

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Use the alternating series test. – Jim Belk Jan 10 '12 at 17:50
I changed "ln \ n" to "\ln n". The backslash not only prevents italicization, but also results in proper spacing between "ln" and "n". Generally with \det, \max, \sup, \sin, etc., the backslash causes the typesetting conventions appropriate to the situation to be followed. For example "\max_{x\in A}" in a "displayed" setting gives you this $\displaystyle\max_{x\in A}$, and in an "inline" setting gives you $\max_{x\in A}$. – Michael Hardy Jan 10 '12 at 19:18

Remark: If you are interested in more then convergence, we can evaluate this explicitly and show that $$\sum_{n=1}^\infty (-1)^n \frac{\log n}{n}=\gamma \log 2 -\frac{(\log 2)^2}{2}$$ where $\gamma$ is the Euler-Mascheroni constant. More generally, we can show that for any $k$ $$\sum_{n=1}^{\infty}\frac{(-1)^{n}\log^{k}n}{n}=\frac{(-1)^{n}\left(\log2\right)^{k+1}}{k+1}+(-1)^{k-1}\sum_{j=0}^{k-1}\gamma_{j}\binom{k}{j}\log^{k-j}2,$$ where the $\gamma_i$ are the Stieltjes constants. See this answer, or this post for specific details regarding these last two facts. (The links are nearly identical)