How to prove that :
$ \sum_{n=1}^{\infty}\frac{1}{2nx^{2n}}=-\frac{1}{2}\ln \left(1-\frac{1}{x^2}\right)$
How to prove that :
$ \sum_{n=1}^{\infty}\frac{1}{2nx^{2n}}=-\frac{1}{2}\ln \left(1-\frac{1}{x^2}\right)$
We factor out $\frac{1}{2}$ as it is a constant $ \sum_{n=1}^{\infty}\frac{1}{2nx^{2n}}= \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{nx^{2n}} $
Now, the Taylor's series for $-\log (1-x)$ is, for $-1\le x < 1$
$-\log (1-x)=\sum^{\infty}_{n=1} \frac{x^n}n$
Thus
$-\log \left(1-\frac{1}{x}\right)=\sum^{\infty}_{n=1} \frac{1}{n x^n}$
$-\log \left(1-\frac{1}{x^2}\right)=\sum^{\infty}_{n=1} \frac{1}{n x^{2n}}$
and, by multiplying both side by $\frac{1}{2}$, we get
$\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{nx^{2n}}=-\log \left(1-\frac{1}{x^2}\right)$