How to evaluate $\sum_{n=1}^\infty \frac{1}{n 2^{2n}}$? 
Evaluate $$\sum_{n=1}^\infty \frac{1}{n2^{2n}}$$

I'd be glad for guidance, because, frankly, I don't have clue here.

For any $z$ such that $|z|<1$ we have:
$$\sum_{n=1}^{+\infty}\frac{z^n}{n} = \sum_{n=1}^{+\infty}\int_{0}^{z}x^{n-1}\,dx = \int_{0}^{z}\frac{dx}{1-x} = -\log(1-z) $$
and by plugging in $z=\frac{1}{4}$ it follows that:
$$ \sum_{n\geq 1}\frac{1}{n 2^{2n}}=\color{red}{\log 4-\log 3}. $$
