Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^22^n}$ How can I prove that 
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2).$$
Can anyone help me please?
 A: \begin{eqnarray}
\sum\limits_{n=1}^\infty \frac{H_n}{n^2 2^n} = \sum\limits_{m=1}^\infty \frac{1}{m} \sum\limits_{n=m}^\infty \frac{1}{n^2 2^n} = \sum\limits_{m=1}^\infty \frac{1}{m} \int\limits_{-\infty}^0(-\xi) \frac{(1/2 \exp(\xi))^m}{1-1/2 \exp(\xi)} d\xi = \\
\int\limits_{-\infty}^0 \xi \frac{\log(1 - 1/2 \exp(\xi))}{1-1/2 \exp(\xi)} d\xi = \\
\int\limits_{1/2}^1 \left(\frac{1}{u} + \frac{1}{1-u}\right) \log(u) \left[\log(2) + \log(1-u)\right] du = \\
\zeta(3) - \frac{1}{12} \pi^2 \log(2)
\end{eqnarray}
I think that all the steps are clear except for the last two ones.In the second last step I substituted for 1 - 1/2 exp(xi).The only non-trivial integrals in here are $\int \log(u)/(1-u) du$ and $\int \log(u) \log(1-u)/(1-u) du$. I compute them now.
The first integral is done by expanding the denominator in a series and integrating term by term.
\begin{equation}
\int\limits_{1/2}^1 \frac{\log(u)}{1-u} du = \sum\limits_{p=0}^\infty \int\limits_{1/2}^1 u^p \log u du = \sum\limits_{p=0}^\infty \frac{-1+2^{-1-p}}{(p+1)^2}  + \log(2) \sum\limits_{p=0}^\infty \frac{2^{-1-p}}{p+1} = -\zeta(2) + Li_2(1/2) + \log(2) Li_1(1/2)
\end{equation}
The second integral is done by integrating by parts and using the definition of the polylogarithmic function.
\begin{equation}
\int\limits_{1/2}^1 \log(u) \frac{\log(1-u)}{1-u} du = \int\limits_{1/2}^1 \log(1-u) Li_2^{'}(1-u) du = \left.\log(1-u) Li_2(1-u)\right|_{1/2}^1 + Li_3(1/2) = \log(2) Li_2(1/2) + Li_3(1/2)
\end{equation}
Now, the only thing that remains is to bring the results together. I am sorry but due to time constraints I am not able to do it right now. I have verified with Mathematica that all the partial results are correct.
Final Note: We can clearly see that the result is expressed though elementary functions and through polylogarithms of order not bigger than three, evaluated at 1/2.From the Wikipedia page on Polylogarithms we learn those polylogarithms at 1/2 are expressed in closed form through $\pi$, $\log(2)$ and the $\zeta$ functions. Having said that we can say that this completes the proof.
A: Starting with  $$ \frac{\ln(1-x)}{1-x}=-\displaystyle \sum_{n=1}^{\infty}H_n x^n $$
multiply both sides by $ \frac{\ln x}{x} $ then integrate from $ x=0 $ to $ 1/2 $ we get
\begin{align*}
 I&= \int_0^{1/2}\frac{\ln x\ln(1-x)}{x(1-x)}\,dx=-\sum_{n=1}^{\infty}H_n \int_0^{1/2} x^{n-1}\ln x\ dx\\
&=-\sum_{n=1}^\infty H_n\left( \frac{\ln2}{2^n n}+\frac{1}{2^n n^2}\right)=-\frac12\ln2\zeta(2)-\sum_{n=1}^\infty \frac{H_n}{2^n n^2}  \tag{1}
\end{align*}
Note that we used $\sum_{n=1}^\infty \frac{H_n}{2^nn}=\frac12\zeta(2)$ which follows from using the generating function $\sum_{n=1}^\infty\frac{x^n H_n}{n}=\operatorname{Li_2}(x)+\frac12 \ln^2(1-x)$ with $x=1/2$ where $\operatorname{Li_2}(1/2)=\frac12\zeta(2)-\frac12\ln^22$

on the other hand
\begin{equation*}
I=  \int_{0}^{1/2}\frac{\ln x\ln(1-x)}{x(1-x)}\ dx \overset{x\mapsto1-x}{=} 
\int_{1/2}^{1}\frac{\ln(1-x) \ln(x)}{x(1-x)}\ dx
\end{equation*}
and by adding the integral to both sides, we get
\begin{equation*}
2I= \int_{0}^{1}\frac{\ln x\ln(1-x)}{x(1-x)}\ dx
=-\sum_{n=1}^\infty H_n \int_0^1 x^{n-1}\ln x \ dx
=\sum_{n=1}^\infty \frac{H_n}{n^2}=2\zeta(3)\tag2
\end{equation*}
where the last result follows from using Euler Identity.
By combining $(1)$ and $(2)$ we obtain the closed form of $\sum_{n=1}^\infty\frac{H_n}{2^nn^2}$.
A: \begin{align*}
\operatorname{Li}_3\left(\frac{1}{2}\right)&=\int _0^{\frac{1}{2}}\frac{\operatorname{Li}_2\left(x\right)}{x}\:dx=-\operatorname{Li}_2\left(\frac{1}{2}\right)\ln \left(2\right)+\int _{\frac{1}{2}}^{1}\frac{\ln \left(x\right)\ln \left(1-x\right)}{1-x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{2}\ln ^3\left(2\right)+\int _0^1\frac{\ln \left(1-x\right)\ln \left(x\right)}{x}\:dx-\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln \left(1-x\right)}{1-x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{2}\ln ^3\left(2\right)+\sum _{k=1}^{\infty }\frac{1}{k^3}-\frac{1}{2}\ln ^3\left(2\right)-\frac{1}{2}\int _0^{\frac{1}{2}}\frac{\ln ^2\left(1-x\right)}{x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)-\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)-\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+\operatorname{Li}_3\left(\frac{1}{2}\right)
\end{align*}
And magically we find the value for that sum
\begin{align*}
\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)
\end{align*}
A: Let's start with the product of $\;-\ln(1-x)\,$ and $\dfrac 1{1-x}$ to get the product generating function
(for $|x|<1$) :
$$\tag{1}f(x):=-\frac {\ln(1-x)}{1-x}=\sum_{n=1}^\infty H_n\, x^n$$
Dividing by $x$ and integrating we get :
\begin{align}
\sum_{n=1}^\infty \frac{H_n}n\, x^n&=\int \frac{f(x)}xdx\\
&=-\int \frac{\ln(1-x)}{1-x}dx-\int\frac{\ln(1-x)}xdx\\
\tag{2}&=C+\frac 12\ln(1-x)^2+\operatorname{Li}_2(x)\\
\end{align}
(with $C=0$ from $x=0$)
The first integral was obtained by integration by parts, the second from the integral definition of the dilogarithm or the recurrence for the polylogarihm (with $\;\operatorname{Li}_1(x)=-\ln(1-x)$) : $$\tag{3}\operatorname{Li}_{s+1}(x)=\int\frac {\operatorname{Li}_{s}(x)}x dx$$
Dividing $(2)$ by $x$ and integrating again returns (using $(3)$ again) :
\begin{align}
\sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n&=\int \frac {\ln(1-x)^2}{2\,x}dx+\int \frac{\operatorname{Li}_2(x)}x dx\\
&=C+I(x)+\operatorname{Li}_3(x)\\
\end{align}
with $I(x)$ obtained by integration by parts (since $\frac d{dx}\operatorname{Li}_2(1-x)=\dfrac {\ln(x)}{1-x}$) :
\begin{align}
I(x)&:=\int \frac {\ln(1-x)^2}{2\,x}dx\\
&=\left.\frac{\ln(1-x)^2\ln(x)}{2}\right|+\int \ln(1-x)\frac {\ln(x)}{1-x}dx\\
&=\left.\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)\right|+\int \frac{\operatorname{Li}_2(1-x)}{1-x}dx\\
&=\left.\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)-\operatorname{Li}_3(1-x)\right|\\
\end{align}
getting the general relation :
$$\tag{4}\sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n=C+\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)$$
(with $C=\operatorname{Li}_3(1)=\zeta(3)$ here)
applied to $x=\dfrac 12$ with $\operatorname{Li}_2\left(\frac 12\right)=\dfrac{\zeta(2)-\ln(2)^2}2$ from the link returns the wished :
\begin{align}
\sum_{n=1}^\infty \frac{H_n}{n^2\;2^n}&=\zeta(3)-\frac{\ln(2)^3}2-\ln(2)\frac{\zeta(2)-\ln(2)^2}2\\
\tag{5}\sum_{n=1}^\infty \frac{H_n}{n^2\;2^n}&=\zeta(3)-\ln(2)\frac{\zeta(2)}2
\end{align}
