Evaluate $\sum\limits_{n=1}^{\infty} (-1)^n \frac{\ln{n}}{n} $ $$\sum\limits_{n=1}^{\infty} (-1)^n \frac{\ln{n}}{n} $$
Hint : $$ x_n = \frac{\ln{2}}{2} + \frac{\ln{3}}{3} + \cdots \frac{\ln{n}}{n} - \frac{\ln^2{2}}{2} $$
Which converges, if we calculate it's limit we should get $\ln2(\gamma-\frac{\ln{2}}{2})$.
I don't understand where this hint comes from and how it helps us solve the series.
 A: 
I thought that it would be instructive to present a straightforward way to evaluate the series of interest using the Euler Maclaurin Summation Formula.  To that end we proceed.


Note that we can write any alternating sum $\sum_{n=1}^{2N}(-1)^na_n$ as 
$$\sum_{n=1}^{2N}(-1)^na_n=2\sum_{n=1}^N a_{2n}-\sum_{n=1}^{2N}a_n\tag1$$
Using $(1)$, we see that
$$\begin{align}
\sum_{n=1}^{2N}(-1)^n \frac{\log(n)}{n}&=2\sum_{n=1}^N \frac{\log(2n)}{2n}-\sum_{n=1}^{2N}\frac{\log(n)}{n}\\\\
&=\log(2)\sum_{n=1}^N\frac1n-\sum_{n=N+1}^{2N}\frac{\log(n)}{n}\tag2
\end{align}$$
Applying the Euler Maclaurin Summation Formula to the second summation on the right-hand side of $(2)$ reveals
$$\begin{align}
\sum_{n=N+1}^{2N}\frac{\log(n)}{n}&=\int_N^{2N}\frac{\log(x)}{x}\,dx+O\left(\frac{\log(N)}{N}\right)\\\\
&=\frac12 \log^2(2N)-\frac12\log^2(N)+O\left(\frac{\log(N)}{N}\right)\\\\
&=\frac12\log^2(2)+\log(2)\log(N)+O\left(\frac{\log(N)}{N}\right)\tag3
\end{align}$$
Substitution of $(3)$ into $(2)$ yields
$$\begin{align}
\sum_{n=1}^{2N}(-1)^n \frac{\log(n)}{n}&=\log(2)\left(-\log(N)+\sum_{n=1}^N \frac1n\right)-\frac12\log^2(2)+O\left(\frac{\log(N)}{N}\right)
\end{align}$$
Finally, using the limit definition of the Euler-Mascheroni constant 
$$\gamma\equiv\lim_{N\to\infty}\left(-\log(N)+\sum_{n=1}^N\frac1n\right)$$
we arrive at the coveted limit 
$$\sum_{n=1}^\infty\frac{(-1)^n\log(n)}{n}=\gamma\log(2)-\frac12\log^2(2)$$
A: Here is another method using analytic regularization.
We have $\eta(s)=\left(1-2^{1-s}\right) \zeta(s)$, and so about $s=1$
$$
\eta(s)=\left(\log(2)(s-1) - \frac{\log^2(2)}{2} \, (s-1)^2 + {\cal O}\left((s-1)^3\right)\right) \zeta(s) \\
\zeta(s) = -\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} {\rm d}\lambda \, \lambda^{-s} \, \frac{\rm d}{{\rm d}\lambda} \log \left( \frac{\sin(\pi\lambda)}{\pi \lambda} \right)
$$
where for $s>1$ the contour can be closed to the right and the residue theorem is used. For regularity at $\lambda=0$, $s<2$ is required also. Substituting $\lambda=it$
$$
\zeta(s) = \frac{\sin\left(\frac{\pi s}{2}\right)}{\pi} \int_{0}^{\infty} {\rm d}t \, t^{-s} \, \frac{\rm d}{{\rm d}t} \log \left( \frac{\sinh(\pi t)}{\pi t} \right) \\
\stackrel{{\rm P.I. | s>1}}{=} \frac{\sin\left(\frac{\pi s}{2}\right)}{\pi (s-1)} \int_{0}^{\infty} {\rm d}t \, t^{1-s} \, \frac{\rm d^2}{{\rm d}t^2} \log \left( \frac{\sinh(\pi t)}{\pi t} \right)
$$
where the second line now converges for $0<s<2$ and hence
$$
\eta(s) = \left(\log(2) - \frac{\log^2(2)}{2} \, (s-1) + {\cal O}\left((s-1)^2\right)\right) \frac{\sin\left(\frac{\pi s}{2}\right)}{\pi} \int_{0}^{\infty} {\rm d}t \, t^{1-s} \, \frac{\rm d^2}{{\rm d}t^2} \log \left( \frac{\sinh(\pi t)}{\pi t} \right) \, .
$$
Deriving with respect to $s$ and setting $s=1$
$$
\eta'(1)=-\frac{\log(2)}{\pi} \int_0^\infty {\rm d}t \log(t) \, \frac{\rm d^2}{{\rm d}t^2} \log \left( \frac{\sinh(\pi t)}{\pi t} \right) - \frac{\log^2(2)}{2\pi} \int_0^\infty {\rm d}t \, \frac{\rm d^2}{{\rm d}t^2} \log \left( \frac{\sinh(\pi t)}{\pi t} \right) \\
=-{\log(2)} \int_0^\infty {\rm d}t \log(t) \, \frac{\rm d}{{\rm d}t} \left( \coth(\pi t) - \frac{1}{\pi t}\right) - \frac{\log^2(2)}{2}
$$
we have
$$
\coth(\pi t)-\frac{1}{\pi t} = \frac{2t}{\pi} \sum_{k=1}^\infty \frac{1}{k^2+t^2} \, .
$$
When interchanging summation and integration order we acquire divergencies, because $\coth(\infty)=1$, but each summand vanishes for $t\rightarrow \infty$. Due to the uniqueness of the result, it does not change up to some divergent part though:
$$
-{\log(2)} \int_0^\infty {\rm d}t \log(t) \, \frac{\rm d}{{\rm d}t} \left( \coth(\pi t) - \frac{1}{\pi t}\right) \\ 
\sim -\log(2) \sum_{k=1}^N \int_0^\infty {\rm d}t \, \log(t) \frac{\rm d}{{\rm d t}} \frac{2t/\pi}{k^2+t^2} \\
=\log(2) \sum_{k=1}^N \int_0^\infty {\rm d}t \,  \frac{2/\pi}{k^2+t^2} \\
=\log(2) \sum_{k=1}^N \frac{1}{k} \\
= \log(2) \left\{ \log(N) + \gamma + {\cal O}(1/N) \right\}
$$
and therefore
$$
\eta'(1)=\gamma \log(2) - \frac{\log^2(2)}{2} \, .
$$
