Evaluating $ \sum\limits_{n=1}^\infty \frac{1}{n^2 2^n} $

Given: $$ \text{S}= \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2.2^n} $$

Consider the following Lemma,

Lemma:

$$ \displaystyle \int_{0}^1 \dfrac{ \ln x }{x+1} \mathrm{d}x = -\dfrac{\pi^2}{12}$$

Proof:

$ \displaystyle \int_{0}^1 \dfrac{ \ln x }{x+1} \mathrm{d}x = \displaystyle \int_{0}^1 \dfrac{ -\ln x }{1-x}\mathrm{d}x + \displaystyle \int_{0}^1 \dfrac{ -2\ln x }{x^2-1}\mathrm{d}x$

Also,

$\displaystyle \int_{0}^1 \dfrac{ -\ln x }{1-x}\mathrm{d}x = \displaystyle \int_{0}^1 \dfrac{ -\ln (1-x) }{x}\mathrm{d}x = \displaystyle \int_{0}^1 \dfrac{x+\frac{x^2}{2}+\frac{x^3}{3}+...}{x}\mathrm{d}x = \sum_{k=1}^\infty \dfrac{1}{k^2} = \frac{\pi^2}{6}$

and,

$\displaystyle \int_{0}^1 \dfrac{ \ln x }{x^2-1}\mathrm{d}x = \displaystyle \sum_{n=0}^{\infty} \int_{0}^1 x^{2n}.{\ln{x}} \ \mathrm{d}x $

$=- \displaystyle \sum_{n=0}^{\infty} {\dfrac{1}{(2n+1)^2}}$ (Using Integration By Parts)

$= \displaystyle \sum_{n=0}^{\infty} {\dfrac{1}{(2n)^2}} - \sum\limits_{n=0}^{\infty} {\dfrac{1}{n^2}}$

$=\dfrac{-3}{4}\zeta(2)= -\dfrac{\pi^2}{8}$

$\therefore \displaystyle \int_{0}^1 \dfrac{ \ln x }{x+1} \mathrm{d}x = \dfrac{\pi^2}{6} - \dfrac{\pi^2}{4} = -\dfrac{\pi^2}{12}$

This completes the proof of our Lemma.

Now,

$ \text{S}= \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2.2^n} $

$= - \displaystyle \sum_{n=1}^{\infty} \left( \dfrac{2^n-1}{n^2.2^n} - \dfrac{1}{n^2} \right)$

$ = - \displaystyle {\ln 2} \sum_{n=1}^{\infty} \left(\dfrac{1}{n} \int_{0}^1 2^{-nx} \mathrm{d}x \right) + \dfrac{\pi^2}{6}$

$ = - \displaystyle {\ln 2} \int_{0}^1 \left(\sum_{n=1}^{\infty} \dfrac{1}{n} \times {2^{-nx}} \right)\mathrm{d}x + \dfrac{\pi^2}{6}$

$ = \displaystyle {\ln 2} \int_{0}^1 \ln \left( \dfrac{2^x - 1}{2^x} \right)\mathrm{d}x + \dfrac{\pi^2}{6}$

$ = \displaystyle {\ln 2} \int_{0}^1 \ln (2^x - 1)\ \mathrm{d}x + \dfrac{\pi^2}{6} -\dfrac{\ln^2 2}{2}$

Now, substituting $(2^x - 1) = t$, we have,

$\text{S} = {\ln 2} \times \dfrac{1}{\ln 2} \displaystyle \int_{0}^1 \dfrac{ \ln t }{t+1} \mathrm{d}t + \dfrac{\pi^2}{6} -\dfrac{\ln^2 2}{2}$

$ = -\dfrac{\pi^2}{12} + \dfrac{\pi^2}{6} -\dfrac{\ln^2 2}{2}$ (Using the Lemma)

$=\boxed{\dfrac{\pi^2}{12} - \dfrac{\ln^2 2}{2}}$

You series equals $\operatorname{Li}_2\left(\frac{1}{2}\right)$. By the functional identity: $$\operatorname{Li}_2(z)+\operatorname{Li}_2(1-z)=\zeta(2)-\log(z)\log(1-z)\tag{1}$$ that is straightforward to prove by differentiation, since: $$ \frac{d}{dz}\operatorname{Li}_2(z) = \frac{d}{dz}\sum_{n\geq 1}\frac{z^n}{n^2}=-\frac{\log(1-z)}{z},\tag{2}$$ we have: $$\sum_{n\geq 1}\frac{1}{n^2 2^n}=\operatorname{Li}_2\left(\frac{1}{2}\right)=\color{red}{\frac{1}{2}\left(\frac{\pi^2}{6}-\log^2 2\right)}.\tag{3}$$

Recall that $$ \sum_{k=1}^\infty\frac{x^k}k=-\log(1-x)\tag{1} $$ Dividing $(1)$ by $x$ and integrating yields $$ \begin{align} \sum_{k=1}^\infty\frac1{2^kk^2} &=-\int_0^{1/2}\frac{\log(1-x)}{x}\,\mathrm{d}x\tag{2a}\\ &=-\left[\vphantom{\int}\log(1-x)\log(x)\right]_0^{1/2} -\int_0^{1/2}\frac{\log(x)}{1-x}\,\mathrm{d}x\tag{2b}\\ &=-\log(2)^2-\int_{1/2}^1\frac{\log(1-x)}x\,\mathrm{d}x\tag{2c}\\ &=-\frac{\log(2)^2}2-\frac12\int_0^1\frac{\log(1-x)}x\,\mathrm{d}x\tag{2d}\\ &=-\frac{\log(2)^2}2+\frac12\sum_{k=1}^\infty\frac1{k^2}\tag{2e}\\ &=\bbox[5px,border:2px solid #00A000]{-\frac{\log(2)^2}2+\frac{\pi^2}{12}}\tag{2f} \end{align} $$ Explanation:
$\text{(2a)}$: divide $(1)$ by $x$ and integrate over $[0,1/2]$
$\text{(2b)}$: integrate by parts
$\text{(2c)}$: substitute $x\mapsto1-x$
$\text{(2d)}$: average $\text{(2a)}$ and $\text{(2c)}$
$\text{(2e)}$: divide $(1)$ by $x$ and integrate over $[0,1]$
$\text{(2f)}$: $\zeta(2)=\frac{\pi^2}6$