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?

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}

We will make frequent use of $$ \binom{n+1}{k+1}=\binom{n}{k}\frac{n+1}{k+1}\tag{1} $$ The Generalized Harmonic Numbers of the second order are defined as $$ H_n^{(2)}=\sum_{k=1}^n\frac1{k^2}\tag{2} $$ The factor of $2^{-n}$ in each term reminded me of the Euler Series Transformation. Reversing the series acceleration (series deceleration?), it can be seen that we should look at $$ \begin{align} \sum_{k=0}^n\binom{n}{k}(-1)^k\frac{H_{k+1}^{(2)}}{k+1} &=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{k+1}\sum_{j=0}^k\frac1{(j+1)^2}\tag{3}\\ &=\frac1{n+1}\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}\sum_{j=0}^k\frac1{(j+1)^2}\tag{4}\\ &=\frac1{n+1}\sum_{j=0}^n\sum_{k=j}^n(-1)^k\binom{n+1}{k+1}\frac1{(j+1)^2}\tag{5}\\ &=\frac1{n+1}\sum_{j=0}^n(-1)^j\binom{n}{j}\frac1{(j+1)^2}\tag{6}\\ &=\frac1{(n+1)^2}\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac1{j+1}\tag{7}\\ &=\frac1{(n+1)^2}\sum_{j=0}^n(-1)^j\sum_{k=0}^n\binom{k}{j}\frac1{j+1}\tag{8}\\ &=\frac1{(n+1)^2}\sum_{k=0}^n\frac1{k+1}\sum_{j=0}^n(-1)^j\binom{k+1}{j+1}\tag{9}\\ &=\frac1{(n+1)^2}\sum_{k=0}^n\frac1{k+1}\tag{10}\\[3pt] &=\frac{H_{n+1}}{(n+1)^2}\tag{11} \end{align} $$ Explanation:
$\ \;(3)$: use $(2)$
$\ \;(4)$: apply $(1)$
$\ \;(5)$: change order of summation
$\ \;(6)$: $\sum\limits_{j=k}^n(-1)^j\binom{n+1}{j+1}=(-1)^k\binom{n}{k}$
$\ \;(7)$: apply $(1)$
$\ \;(8)$: $\sum\limits_{j=0}^n\binom{j}{k}=\binom{n+1}{k+1}$
$\ \;(9)$: $\sum\limits_{j=0}^k(-1)^j\binom{k+1}{j+1}=1$
$(10)$: $H_{n+1}=\sum\limits_{k=0}^n\frac1{k+1}$

Using $(11)$, the Euler Series Transformation says that $$ \begin{align} \sum_{k=0}^\infty(-1)^k\frac{H_{k+1}^{(2)}}{k+1} &=\sum_{n=0}^\infty2^{-n-1}\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{H_{k+1}^{(2)}}{k+1}\\ &=\sum_{n=0}^\infty2^{-n-1}\frac{H_{n+1}}{(n+1)^2}\tag{12} \end{align} $$ Therefore, with a change of indexing, we get $$ \begin{align} \sum_{n=1}^\infty2^{-n}\frac{H_n}{n^2} &=\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^{(2)}}{n}\tag{13}\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^3}+\sum_{n=1}^\infty(-1)^{n-1}\frac{H_{n-1}^{(2)}}{n}\tag{14}\\ &=\frac34\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\sum_{k=1}^{n-1}\frac1{k^2}\tag{15}\\ &=\frac34\zeta(3)+\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac{(-1)^{n-1}}{nk^2}\tag{16}\\ &=\frac34\zeta(3)+\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{k+n-1}}{(k+n)k^2}\tag{17}\\ &=\frac34\zeta(3)+\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{k+n-1}\left(\frac1{k^2n}-\frac1{kn(k+n)}\right)\tag{18}\\[6pt] &=\frac34\zeta(3)-\frac12\zeta(2)\log(2)+\frac14\zeta(3)\tag{19}\\[9pt] &=\zeta(3)-\frac12\zeta(2)\log(2)\tag{20} \end{align} $$ Explanation:
$(13)$: reindex $(12)$
$(14)$: $H_n^{(2)}=\frac1{n^3}+H_{n-1}^{(2)}$
$(15)$: apply $(2)$
$(16)$: change order of summation
$(17)$: reindex $n\mapsto k+n$
$(18)$: $\frac1{(k+n)k^2}=\frac1{k^2n}-\frac1{kn(k+n)}$
$(19)$: $\sum\limits_{k=1}^\infty\sum\limits_{n=1}^\infty\frac{(-1)^{k+n}}{kn(k+n)}=\frac14\zeta(3)$ from $(5)$ and $(7)$ of this answer
$(20)$: addition

\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.

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 $\sum_{n=1}^\infty\frac{H_n}{2^nn^2}$.