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 could write this just as well as
$$
\sum_{n=1}^\infty \frac{1}{n}(1/4)^n
$$
Consider the function
$$
f(x) = \sum_{n=1}^\infty \frac{1}{n}x^n \quad x \in (-1,1)
$$
noting that $f(0) = 0$.
We evaluate
$$
f'(x) = \sum_{n=0}^\infty x^n = \frac{1}{1-x} \quad x \in (-1,1)
$$
It follows that
$$
f(4) = f(0) + \int_0^{1/4} f'(t)\,dt = \int_0^{1/4} \frac 1{1-t}\,dt
$$
