TITLE: $\sum_{n\geq 1}\frac{(-1)^n \ln n}{n}$
QUESTION [8 upvotes]: How can we compute the series $\displaystyle \sum_{n\geq 1}\frac{(-1)^n \ln n}{n}$?
I know it is $\eta '(1)$ , where $\eta$ is the $\eta$ Dirichlet Function , i know its value. But I don't know how to compute it. 
An approach I tried is to  expand the series, then gather together the odd  and the even terms  , use $\zeta$ (Riemann's function) and that's all. Then no idea.
Any ideas are welcome.

REPLY [11 votes]: This is given in equation $(8)$ of this answer using the Euler-Maclaurin Sum Formula, which says that
$$
\sum_{k=1}^n\frac{\log(k)}{k}=C+\frac{\log(n)^2}{2}+O\left(\frac{\log(n)}{n}\right)\tag{1}
$$
for some constant $C$, and that
$$
\sum_{k=1}^n\frac1{k}=\log(n)+\gamma+O\left(\frac1n\right)\tag{2}
$$
Note that
$$
\begin{align}
\sum_{k=1}^{2n}(-1)^k\frac{\log(k)}{k}
&=2\sum_{k=1}^{n}\frac{\log(2k)}{2k}
-\sum_{k=1}^{2n}\frac{\log(k)}{k}\\
&=\sum_{k=1}^{n}\frac{\log(k)}{k}+\log(2)\sum_{k=1}^{n}\frac1{k}
-\sum_{k=1}^{2n}\frac{\log(k)}{k}\\
&=C+\frac{\log(n)^2}{2}+O\left(\frac{\log(n)}{n}\right)\\
&+\log(2)\left(\log(n)+\gamma+O\left(\frac1n\right)\right)\\
&-C-\frac{\log(2n)^2}{2}-O\left(\frac{\log(2n)}{2n}\right)\\
&=\gamma\log(2)-\frac{\log(2)^2}{2}+O\left(\frac{\log(n)}{n}\right)\tag{3}
\end{align}
$$
Let $n\to\infty$ in $(3)$ and we get
$$
\sum_{k=1}^\infty(-1)^k\frac{\log(k)}{k}=\gamma\log(2)-\frac{\log(2)^2}{2}\tag{4}
$$
Note that it doesn't matter what the constant $C$ is, as it gets cancelled out of the calculations.