$\sum_{n\geq 1}\frac{(-1)^n \ln n}{n}$ 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.
 A: Let $s>1$. Observe that $$\displaystyle \sum_{n\geq 1}\frac{(-1)^n }{n^s}=\frac{1}{2^s}\sum_{n\geq 1}\frac{1}{n^s}-\sum_{n\geq 0}\frac{1}{(2n+1)^s}\tag1$$
and
$$\displaystyle \sum_{n\geq 1}\frac{1}{n^s}=\frac{1}{2^s}\sum_{n\geq 1}\frac{1}{n^s}+\sum_{n\geq 0}\frac{1}{(2n+1)^s}\tag2$$
then $(1)$ + $(2)$ gives
$$\displaystyle \sum_{n\geq 1}\frac{(-1)^n }{n^s}=\left(\frac{1}{2^{s-1}}-1\right)\sum_{n\geq 1}\frac{1}{n^s}=\left(\frac{1}{2^{s-1}}-1\right)\zeta(s)\tag3$$
where $\zeta(\cdot)$ is the Riemann zeta function.
Now differentiating $(3)$ with respect to $s$ gives 
$$\displaystyle -\sum_{n\geq 1}\frac{(-1)^n \ln n}{n^s}=\left(-\frac{\ln 2}{2^{s-1}}\right)\zeta(s)+\left(\frac{1}{2^{s-1}}-1\right)\zeta'(s)\tag4$$
 and letting $s$ tend to $1^+$, gives the result:

$$\displaystyle \sum_{n\geq 1}\frac{(-1)^n }{n}\ln n=-\frac{\ln^2 2}{2}+\gamma \ln 2\tag5$$

where $\gamma$ is the Euler constant and where, near $s=1^+$, we have used 
$$\frac{1}{2^{s-1}}= 1- (s-1)\ln 2+(s-1)^2\frac{\ln^2 2}{2}+\mathcal{O}((s-1)^3)$$
together with the Laurent series expansions $$
\begin{align}
\zeta(s) &=\frac{1}{s-1}+\gamma+\mathcal{O}(s-1)\\\\
\zeta'(s)&=-\frac{1}{(s-1)^2}+\mathcal{O}(1).\end{align}$$
A: 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.
