The value of $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\log n}{n}$ How to compute the following convergent series? or some hint!
$$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\log n}{n}.$$
The Wolfram MATHEMATICA9.0 gives the result is $1/2(\log 2)^2-\gamma\log 2$,
where $\gamma$ is tha Euler constant!
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$\ds{\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{\ln\pars{n} \over n}:\
{\large ?}}$.

Hereafter, I'll use
well known identities which are related to the Riemann Zeta Function:
\begin{align}
&\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{\ln\pars{n} \over n} =
\left.\partiald{}{\mu}\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n^{1 - \mu}}
\right\vert_{\ \mu\ =\ 0^{+}} =
\left.\partiald{\bracks{\pars{1 - 2^{\mu}}\zeta\pars{1 - \mu}}}{\mu}\,
\right\vert_{\ \mu\ =\ 0^{+}}
\\[5mm] = &\
\partiald{}{\mu}
\braces{\bracks{-\ln\pars{2}\mu - {1 \over 2}\,\ln^{2}\pars{2}\,\mu^{2} +
\,\mrm{O}\pars{\mu^{3}}}\bracks{-\,{1 \over \mu} + \gamma + \,\mrm{O}\pars{\mu^{1}}}}_{\ \mu\ =\ 0^{+}}
\\[5mm] = &\
\partiald{}{\mu}\braces{\ln\pars{2} +
\bracks{{1 \over 2}\,\ln^{2}\pars{2} - \gamma\ln\pars{2}}\mu + \,\mrm{O}\pars{\mu^{2}}}_{\ \mu\ =\ 0^{+}}
\\[5mm] = &\
\bbx{{1 \over 2}\,\ln^{2}\pars{2} - \gamma\ln\pars{2}}
\end{align}
A: We can write
$$\begin{align}
\sum_{n=1}^{2N}(-1)^{n-1}\frac{\log(n)}{n}&=\sum_{n=1}^{2N} \frac{\log(n)}{n}-2\sum_{n=1}^N\frac{\log(2n)}{2n}\\\\
&=\sum_{n=N+1}^{2N}\frac{\log(n)}{n}-\log(2)\sum_{n=1}^N \frac1n
\end{align}$$
Using the Euler-Maclaurin Summation Formula, we have
$$\begin{align}
\sum_{n=}^{2N}(-1)^{n-1}\frac{\log(n)}{n}&=\int_{N+1}^{2N}\frac{\log(x)}{x}\,dx-\log(2)\,H_n+O\left(\frac{\log(N)}{N}\right)\\\\
&=\frac12\log^2(2N)-\frac12\log^2(N+1)-\log(2)\log(N)-\log(2)\gamma+O\left(\frac{\log(N)}{N}\right)\\\\
&=\frac12 \log^2(2)-\log(2)\gamma+O\left(\frac{\log(N)}{N}\right)\\\\
\end{align}$$
Taking the limit as $N\to \infty$ yields the coveted limit
$$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\log(n)}{n}=\frac12\log^2(2)-\log(2)\gamma$$
