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.
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.
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 ...
$ \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.
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...