Evaluating an infinite series $\sum_{n=1}^{\infty} \frac{1}{n^{2}2^{n}}$

Hint Let $f(x)=\sum_{n=1}^\infty \frac{1}{n^2}x^n$. Then $$f'(x)=\sum_{n=1}^\infty \frac{1}{n}x^{n-1} \\ xf'(x)=\sum_{n=1}^\infty \frac{1}{n}x^{n}\\ f'(x)+xf''(x)=\sum_{n=1}^\infty x^{n-1}=\frac{1}{1-x}$$

For simplicity let $y=f'(x)$. Then you need to solve: $$y+xy'=\frac{1}{1-x}$$ or equivalently $$(xy)'=\frac{1}{1-x}$$

You can find $y=f'$ and hence $f$ from here. Plug in $x=\frac{1}{2}$.