How evaluate $ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$ How prove
$$ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}=\frac{\pi^2\ln2}{6}-\frac{\ln^32}{3}-\frac{3\zeta(3)}{4} $$
$\mathbf {My\,Attempt:}$
I put $$\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right) = \int_0^1 \frac{x^n}{1+x}dx$$ and hence
$$\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^2=\int_0^1\int_0^1 \frac{(xy)^n}{(1+x)(1+y)}dxdy$$
So, the sum equals $$\int_0^1\int_0^1 \frac{1}{(1+x)(1+y)}\left(\sum\limits_{n=1}^{\infty} \frac{(xy)^n}{n}\right)dxdy=-\int_0^1\int_0^1 \frac{\ln(1-xy)}{(1+x)(1+y)}dxdy$$
The inner integral is
$$\int_0^1 \frac{\ln(1-xy)}{1+y}dy=\mathrm{Li}_2 \left(\frac{1-x}{1+x}\right)-\mathrm{Li}_2\left(\frac{1}{1+x}\right)+\ln(1-x)\ln\left(\frac{2x}{1+x}\right)$$
The remaining stuff is a lot of calculations.  
$\text{Any hint for a better method or idea?}$
 A: The second round of integration is completed with use of 
\begin{align}
I_{1} &= \int_{0}^{1} \mathrm{Li}_2 \left(\frac{1-x}{1+x}\right) \, \frac{dx}{1+x} = \zeta(2) \, \ln 2 - \frac{5}{8} \, \zeta(3) \\
I_{2} &= \int_{0}^{1} \mathrm{Li}_2 \left(\frac{1}{1+x}\right) \, \frac{dx}{1+x} = \frac{\zeta(3)}{8} + \frac{\zeta(2) \, \ln 2}{2} - \frac{\ln^{3}2}{6} \\
I_{3} &= \int_{0}^{1} \ln(1-x) \, \ln\left(\frac{2 x}{1+x}\right) \,  \frac{dx}{1+x} =
\frac{3 \, \zeta(3)}{2} + \frac{\ln^{3}2}{6} - \frac{3}{2} \, \zeta(2) \, \ln 2.
\end{align}
With these values then:
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{n} \, \left( \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k+n} \right)^{2} \\
&= \sum_{n=1}^{\infty} \frac{1}{n} \, \int_{0}^{1} \int_{0}^{1} \frac{(u t)^{n} \, du \, dt}{(1+u) (1+t)} \\
&= - \int_{0}^{1} \int_{0}^{1} \frac{\ln(1- u t) \, du \, dt}{(1+u)(1+t)} \\
&= - \int_{0}^{1} \left( \mathrm{Li}_2 \left(\frac{1-t}{1+t}\right)-\mathrm{Li}_2\left(\frac{1}{1+t}\right)+\ln(1-t)\ln\left(\frac{2t}{1+t}\right) \right) \, \frac{dt}{1+t} \\
&= - \left[ \left(\zeta(2) \, \ln 2 - \frac{5}{8} \, \zeta(3) \right) - \left(\frac{\zeta(3)}{8} + \frac{\zeta(2) \, \ln 2}{2} - \frac{\ln^{3}2}{6} \right) \right. \\
& \hspace{20mm} \left. + \left(\frac{3 \, \zeta(3)}{2} + \frac{\ln^{3}2}{6} - \frac{3}{2} \, \zeta(2) \, \ln 2 \right) \right] \\
&= \zeta(2) \, \ln 2 - \frac{3 \, \zeta(3)}{4} - \frac{\ln^{3}2}{3}.
\end{align}
This leads to saying
$$ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}= \zeta(2)\ln 2 -\frac{3\zeta(3)}{4} - \frac{\ln^{3}2}{3}.$$
A: $$\iint_{(0,1)^2}\frac{-\log(1-xy)}{(1+x)(1+y)}\,dx\,dy = 2\iint_{0\leq y\leq x\leq 1}\frac{-\log(1-xy)}{(1+x)(1+y)}\,dx\,dy $$
equals
$$ 2\int_{0}^{1}\int_{0}^{1}\frac{-x\log(1-x^2 z)}{(1+x)(1+xz)}\,dx\,dz = 2\int_{0}^{1}\frac{\text{Li}_2\left(\frac{1}{1+x}\right)-\text{Li}_2(1-x)-\log(x)\log(1-x^2)}{1+x}\,dx$$
and we may tackle three separate integrals. 
$$ \int_{0}^{1}\text{Li}_2\left(\frac{1}{1+x}\right)\frac{dx}{1+x}=\int_{1/2}^{1}\frac{\text{Li}_2(x)}{x}\,dx=\int_{1/2}^{1}\sum_{n\geq 1}\frac{x^{n-1}}{n^2}\,dx=\zeta(3)-\text{Li}_3\left(\tfrac{1}{2}\right), $$
$$ \int_{0}^{1}\frac{\text{Li}_2(1-x)}{1+x}\,dx = \int_{0}^{1}\frac{\text{Li}_2(x)}{2-x}\,dx\stackrel{\text{IBP}}{=}-\int_{0}^{1}\frac{\log(1-x)\log(2-x)}{x}\,dx=\frac{\pi^2}{4}\log(2)-\zeta(3)$$
and the third one is similar. Interestingly, the second integral equals
$$ \sum_{n\geq 1}\frac{1}{n}\left[\frac{1}{(n+1)^2}+\ldots+\frac{1}{(2n)^2}\right]=\frac{\pi^2}{4}\log(2)-\zeta(3).$$
