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.

We have
$$\sum_{n = 1}^\infty \frac{1}{n2^{2n}} = \sum_{n = 1}^\infty \frac{(1/4)^n}{n} = -\log\left(1 - \frac{1}{4}\right) = \log\frac{4}{3}$$
