TITLE: Calculate sum of $\sum_{n=1}^{\infty}(-1)^n\frac{\ln n}{n}$
QUESTION [13 upvotes]: What is the exact sum of $$\sum_{n=1}^{\infty}(-1)^n\frac{\ln n}{n}$$

REPLY [18 votes]: $$\begin{eqnarray}\sum_{k=1}^{2n}(-1)^k\frac{\log(k)}{k} &=& -\sum_{k=1}^{2n}\frac{\log(k)}{k} + \sum_{k=1}^n \frac{\log(2k)}{k}\\
&=& \log(2)\sum_{k=1}^n\frac{1}{k} -\sum_{k=1}^n \frac{\log(n+k)}{n+k}\\
&=& \log(2)\sum_{k=1}^n\frac{1}{k} -\frac{\log(n)}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}- \frac{1}{n}\sum_{k=1}^n\frac{\log(1+\frac{k}{n})}{1+\frac{k}{n}}\\
\end{eqnarray}$$
Now we can recognize two Riemann sums:
$$\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}=\int_1^2\frac{dt}{t} + O(\frac{1}{n}) = \log(2)+O(\frac{1}{n})$$
and
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{\log(1+\frac{k}{n})}{1+\frac{k}{n}} =\int_1^2\frac{\log(t)}{t}dt = \frac{1}{2}\log(2)^2.$$
Combining all this gives
$$\begin{eqnarray}
\sum_{k=1}^{\infty}(-1)^k\frac{\log(k)}{k}&=&-\frac{1}{2}\log(2)^2+\log(2) \lim_{n\to\infty}\left( \sum_{k=1}^n\frac{1}{k} -\log(n)\right)\\
&=& -\frac{1}{2}\log(2)^2+\log(2)\,\gamma.
\end{eqnarray}$$