Summation of Infinite Series $\sum_{n=1}^{\infty} \frac{1}{n 2^{2n+1}}$ Show That :
$$\sum_{n=1}^{\infty} \frac{1}{n 2^{2n+1}} = \ln \left(\frac{2}{\sqrt{3}}\right)$$
I could show convergence. (I dont need to show that this converges). However I couldn't figure how to show the value.
 A: There are various options - here is one: I won't work it through to the end.
Write $$f(t)=\sum_{n=1}^{\infty} \frac{t^n}{n 2^{2n+1}}$$
Then $$f'(t)=\sum_{n=1}^{\infty} \frac{t^{n-1}}{2^{2n+1}} = \frac 1 8 \sum_{n=1}^{\infty} \left(\frac t 4\right)^{n-1}$$
Which is a geometric series ...
A: $$
\begin{align*}
\sum_{n=1}^{\infty} \frac{y^n}{n} &= -\ln \left(1-y\right) \hspace{15pt} {\textit{apply }} \hspace{5pt} y=\frac{1}{x^2}\\
\sum_{n=1}^{\infty} \frac{1}{n x^{2n}} &= -\ln \left(1-\frac{1}{x^2}\right) \\
\sum_{n=1}^{\infty} \frac{1}{n 2^{2n+1}} &= \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n 2^{2n}} = -\frac{1}{2}\ln \left(1-\frac{1}{4}\right)\\
\end{align*}
$$
Do the rest to simplify and get what you want.
A: Artin's comment pretty much answers the question.

Did you try $\sum_{n=1}^\infty\frac{y^n}{n}=-ln(1-y)$

Just observe that your series actually is $$\frac{1}{2}\sum_{n=1}^\infty\frac{\left(1/4\right)^n}{n}$$
As this is homework, you should also, I guess, be careful to check radius of convergence...
