Prove $\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\frac{\pi ^{2}}{48}+\frac{1}{4}\ln^22$. How to prove the following series,
$$\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\frac{\pi ^{2}}{48}+\frac{1}{4}\ln^22$$
I know a formula which might be usful.
$$\sum_{j=1}^{n}\frac{\left ( -1 \right )^{j-1}}{j}=\ln 2+\left ( -1 \right )^{n-1}\int_{0}^{1}\frac{x^{n}}{1+x}\mathrm{d}x$$
any hint will be appreciate.
 A: Yes! As you mentioned, it's a very useful formula.
Use 
$$\sum_{j=1}^{n}\frac{\left ( -1 \right )^{j-1}}{j}=\ln 2+\left ( -1 \right )^{n-1}\int_{0}^{1}\frac{x^{n}}{1+x}\, \mathrm{d}x$$
we get
\begin{align*}
&\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\ln 2\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}-\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\int_{0}^{1}\frac{x^{2k}}{1+x}\, \mathrm{d}x\\
&=\ln^22+\int_{0}^{1}\frac{1}{1+x}\sum_{k=1}^{\infty }\frac{\left ( -x^{2} \right )^{k}}{k}\, \mathrm{d}x=\ln^22-\int_{0}^{1}\frac{\ln\left ( 1+x^{2} \right )}{1+x}\, \mathrm{d}x
\end{align*}
let
$$f\left ( t \right )=\int_{0}^{1}\frac{\ln\left ( 1+tx^{2} \right )}{1+x}\,\mathrm{d}x$$
then 
\begin{align*}
 f{}'\left ( t \right ) &=\int_{0}^{1}\frac{x^{2}}{\left ( 1+x \right )\left ( 1+tx^{2} \right )}\,\mathrm{d}x \\
 &=\frac{1}{t+1}\int_{0}^{1}\frac{x-1}{1+tx^{2}}\,\mathrm{d}x+\frac{1}{t+1}\int_{0}^{1}\frac{\mathrm{d}x}{x+1} \\
 &=\frac{1}{t+1}\left [ \frac{1}{2t}\ln\left ( 1+tx^{2} \right )-\frac{1}{\sqrt{t}}\arctan\left ( \sqrt{t}x \right )+\ln\left ( x+1 \right ) \right ]_{0}^{1} \\
 &=\frac{1}{t+1}\left [ \frac{1}{2t}\ln\left ( 1+t \right )-\frac{1}{\sqrt{t}}\arctan\left ( \sqrt{t} \right )+\ln 2 \right ]
\end{align*}
Integrate back
$$\Rightarrow f\left ( t \right )=\frac{1}{2}\left [ -\mathrm{Li}_{2}\left ( -t \right )-\frac{1}{2}\ln^{2}\left ( t+1 \right ) \right ]-\arctan^{2}\sqrt{t}+\ln 2\ln\left ( t+1 \right )$$
then let $t=1$ and use $\displaystyle \mathrm{Li}_{2}\left ( -1 \right )=-\frac{\pi ^{2}}{12}$, you will get the answer as wanted.
EDIT:
$$\mathrm{Li}_{2}\left ( -1 \right )=\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k^{2}}=\sum_{k=1}^{\infty }\frac{1}{k^{2}}-2\sum_{k=1}^{\infty }\frac{1}{\left ( 2n-1 \right )^{2}}=\frac{\pi ^{2}}{6}-2\cdot \frac{\pi ^{2}}{8}=-\frac{\pi ^{2}}{12}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{\sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k}
\sum_{j = 1}^{2k}{\pars{-1}^{\,j} \over j}:\ {\large ?}}$.

Note that
\begin{align}
&\sum_{j = 1}^{2k}{\pars{-1}^{\,j} \over j} =
\sum_{j = 1}^{k}{1 \over 2j} - \sum_{j = 1}^{k}{1 \over 2j - 1} =
{1 \over 2}\sum_{j = 1}^{k}{1 \over j} -
\pars{\sum_{j = 1}^{2k}{1 \over j} - \sum_{j = 1}^{k}{1 \over 2j}} =
\sum_{j = 1}^{k}{1 \over j} - \sum_{j = 1}^{2k}{1 \over j}
\\[5mm] = &\ \bbx{\ds{H_{k} - H_{2k}}}\qquad
\pars{~H_{n}:\ Harmonic\ Number~}
\end{align}

Then,
\begin{align}
&\sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k}
\sum_{j = 1}^{2k}{\pars{-1}^{\,j} \over j} =
\sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k}\pars{H_{k} - H_{2k}} =
\sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k}H_{k} -
2\sum_{k = 1}^{\infty }{\ic^{2k} \over 2k}H_{2k}
\\[5mm] = &\
\sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k}H_{k} -
2\sum_{k = 1}^{\infty }{\ic^{k} \over k}H_{k}\,{1 + \pars{-1}^{k} \over 2} =
\sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k}H_{k} -
2\,\Re\sum_{k = 1}^{\infty }{\ic^{k} \over k}H_{k}
\\[5mm] = &\
\,\mrm{f}\pars{-1} - 2\,\Re\pars{\mrm{f}\pars{\ic}}\quad
\mbox{where}\quad\mrm{f}\pars{x} \equiv
\sum_{k = 1}^{\infty}{x^{k} \over k}\,H_{k}\label{1}\tag{1}
\end{align}

However,
\begin{align}
\mrm{f}\pars{x} & =
\sum_{k = 1}^{\infty}x^{k}\,H_{k}\int_{0}^{1}t^{k - 1}\,\dd t =
\int_{0}^{1}{1 \over t}\sum_{k = 1}^{\infty}\pars{xt}^{k}\,H_{k}\,\dd t =
\int_{0}^{1}{1 \over t}\bracks{-\,{\ln\pars{1 - xt} \over 1 - xt}}\,\dd t
\\[5mm] & =
-\int_{0}^{x}{\ln\pars{1 - t} \over t\pars{1 - t}}\,\dd t =
-\int_{0}^{x}{\ln\pars{1 - t} \over t}\,\dd t -
\int_{0}^{x}{\ln\pars{1 - t} \over 1 - t}\,\dd t
\\[5mm] & =
\int_{0}^{x}\mrm{Li}_{2}'\pars{t}\,\dd t +
\bracks{{1 \over 2}\,\ln^{2}\pars{1 - t}}_{\ 0}^{\ x} =
\bbx{\ds{\mrm{Li}_{2}\pars{x}} + {1 \over 2}\,\ln^{2}\pars{1 - x}}
\end{align}

With expression \eqref{1}:
\begin{align}
&\sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k}
\sum_{j = 1}^{2k}{\pars{-1}^{\,j} \over j} =
\mrm{Li}_{2}\pars{-1} + {1 \over 2}\,\ln^{2}\pars{2}
-2\,\Re\pars{\mrm{Li}_{2}\pars{\ic}} - \Re\pars{\ln^{2}\pars{1 - \ic}}
\\[5mm] = &
\bbx{\ds{{\pi^{2} \over 48} + {1 \over 4}\,\ln^{2}\pars{2}}}
\end{align}


Note that
  $\ds{\mrm{Li}_{2}\pars{-1} = -\,{\pi^{2} \over 12}}$,
  $\ds{\Re\pars{\mrm{Li}_{2}\pars{\ic}} = -\,{\pi^{2} \over 48}}$ and
  $$
\Re\pars{\ln^{2}\pars{1 - \ic}} =
\Re\pars{\bracks{{1 \over 2}\,\ln\pars{2} - {\pi \over 4}\,\ic}^{2}} =
{1 \over 4}\,\ln^{2}\pars{2} - {\pi^{2} \over 16}
$$

