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.
 A: $$\sum_{n=1}^\infty \frac{1}{n2^{2n}}=\sum_{n=1}^\infty \frac{(1/4)^n}{n}=-\ln\left(1-\frac14\right)=-\ln\frac34$$

Since,
$$\ln(1+x)=x-x^2/2+x^3/3-x^4/4+...$$
A: 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
$$
f(4) = f(0) + \int_0^{1/4} f'(t)\,dt = \int_0^{1/4} \frac 1{1-t}\,dt
$$
A: We have
$$\sum_{n = 1}^\infty \frac{1}{n2^{2n}} = \sum_{n = 1}^\infty \frac{(1/4)^n}{n} = -\log\left(1 - \frac{1}{4}\right) = \log\frac{4}{3}$$
A: 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:
$$ \sum_{n\geq 1}\frac{1}{n 2^{2n}}=\color{red}{\log 4-\log 3}. $$
A: Think about $\sum_{n=1}^\infty\frac{x^n}{n}=\log{(1-x)}$. The sum of your series is therefore $\log{(4/3)}$
