<s>[INST] 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.
 [/INST] 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:
$$\mathrm{Li}_{2}\left ( -1 \right )=\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k^{2}}=\sum_{k=1}^{\infty }\frac{1}{k^{2}}-2\sum_{k=1}^{\infty }\frac{1}{\left ( 2n-1 \right )^{2}}=\frac{\pi ^{2}}{6}-2\cdot \frac{\pi ^{2}}{8}=-\frac{\pi ^{2}}{12}$$
</s>