How to evaluate $ \sum\limits_{n=1}^{\infty} \left( \frac{H_{n}}{(n+1)^2.2^n} \right)$ 
Evaluate
$$ \sum_{n=1}^{\infty} \left( \dfrac{H_{n}}{(n+1)^2.2^n} \right)$$
Where $H_{n}$ is the $n^{th}$ Harmonic Number, i.e., $H_{n} = \displaystyle \sum _{k=1}^n \frac{1}{k}$

I tried to use the Integral Representation for the Harmonic number i.e.,
$$ H_{n} = \int_{0}^1 \dfrac {1-x^n}{1-x} \mathrm{d}x $$
and then interchanging the summation and integral signs, but it further complicated the problem. I also tried to use a result from my previous problem, i.e.,
$$ \sum_{n=1}^{\infty} \dfrac{1}{n^2.2^n} = \dfrac{\pi^2}{12} - \dfrac{\ln^2 2}{2} $$
but no significant progress so far.
Any help will be appreciated.
Thanks!
 A: We have$$\sum_{n\geq1}H_{n}x^{n}=-\frac{\log\left(1-x\right)}{1-x}$$
 then if we integrate $$\sum_{n\geq1}\frac{H_{n}}{n+1}x^{n+1}=\frac{1}{2}\log^{2}\left(1-x\right)$$
 (note we can take $C=0$ from $x=0$) then$$\sum_{n\geq1}\frac{H_{n}}{n+1}x^{n}=\frac{\log^{2}\left(1-x\right)}{2x}$$
 and if we integrate again $$\sum_{n\geq1}\frac{H_{n}}{\left(n+1\right)^{2}}x^{n+1}=\int\frac{\log^{2}\left(1-x\right)}{2x}dx.$$
 Now if we integrate by parts$$\int\frac{\log^{2}\left(1-x\right)}{2x}dx=\frac{\log^{2}\left(1-x\right)\log\left(x\right)}{2}+\int\frac{\log\left(1-x\right)\log\left(x\right)}{1-x}dx$$
 and using the facts $\textrm{Li}'_{2}\left(1-x\right)=\frac{\log\left(x\right)}{1-x}$
  and $\textrm{Li}'_{3}\left(1-x\right)=-\frac{\textrm{Li}{}_{2}\left(1-x\right)}{1-x}$
  we have $$\int\frac{\log^{2}\left(1-x\right)}{2x}dx=\frac{\log^{2}\left(1-x\right)\log\left(x\right)}{2}+\log\left(x\right)\textrm{Li}{}_{2}\left(1-x\right)-\textrm{Li}{}_{3}\left(1-x\right)+C.$$
 If we put $x=1$ we get $$\sum_{n\geq1}\frac{H_{n}}{\left(n+1\right)^{2}}=s_{h}\left(1,2\right)$$ which is an Euler's sum, an it can be calculated by the formula $$\sum_{n\geq1}\frac{H_{n}}{\left(n+1\right)^{m}}=\sum_{n\geq1}\frac{H_{n}}{n^{m}}-\zeta(1+m)$$ (note we have the computation $\sum_{n\geq1}\frac{H_{n}}{n^{2}}=2\zeta(3))$. So we can take $C=\zeta(3)$, then $$\sum_{n\geq1}\frac{H_{n}}{\left(n+1\right)^{2}}x^{n}=\frac{1}{x}\left(\frac{\log\left(1-x\right)\log\left(x\right)}{2}+\log\left(x\right)\textrm{Li}_{2}\left(1-x\right)-\textrm{Li}_{3}\left(1-x\right)+\zeta\left(3\right)\right).$$
A: On the internet you can find that:
$$ \sum_{n=1}^{\infty} \left( \dfrac{H_{n}}{n^2.2^n} \right)=\zeta(3)-\frac{\pi^2ln(2)}{12}$$
Now if we rearrange the sum  :
$$\begin{align}\sum_{n=1}^{\infty} \left( \dfrac{H_{n}}{(n+1)^2.2^n} \right)&=2.\sum_{n=1}^{\infty} \left( \dfrac{H_{n}}{(n+1)^2.2^{n+1}} \right)\\
&=2\left(\zeta(3)-\frac{\pi^2ln(2)}{12}-\frac{1}{2}-\sum_{i=1}^{+\infty}\frac{1}{(n+1)^32^n}\right)\end{align}$$
and it's also known that :
$$\sum_{i=1}^{+\infty}\frac{1}{n^32^n}=\frac{ln(2)^3}{6}-\frac{\pi^2ln(2)}{12}+\frac{7}{8}\zeta(3) $$
now it's your turn to rearrange all to get the result!all this sums can be found as comments in this related post.
