real analysis - Infinite Series $sum_{n=1}^inftyfrac{H_n}{n^22^n}$

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?

Answer

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}