<s>[INST] Evaluating $ \sum\limits_{n=1}^\infty \frac{1}{n^2 2^n} $ 
Evaluate
$$ \sum_{n=1}^\infty \dfrac{1}{n^2 2^n}. $$

I have tried using the Maclaurin series of $2^{-n}$ but it further complicated the question. Moreover, I have also tried taking help from another question when the $n^2$ is in the numerator, but no significant progress so far.
Any help will be appreciated.
Thanks!
 [/INST] the taylor series of $$y=\frac{1}{x}\log(\frac{1}{1-x})$$ is
$$y=1+\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+...\frac{x^{n-1}}{n}$$
$$\int_{0}^{0.5}ydx=x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+\frac{x^4}{4^2}+...\frac{x^n}{n^2}=\sum_{n=1}^{\infty }\frac{(.5)^n}{n^2}=\sum_{n=1}^{\infty }\frac{1}{2^nn^2}$$
$$\sum_{n=1}^{\infty }\frac{1}{2^nn^2}=\int_{0}^{0.5}\frac{1}{x}\log\frac{1}{1-x}dx$$
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