<s>[INST] Proof : $ \sum_{n=1}^{\infty}\frac{1}{2nx^{2n}}=-\frac{1}{2}\ln (1-\frac{1}{x^2})$ How to prove that :
$$ \sum_{n=1}^{\infty}\frac{1}{2nx^{2n}}=-\frac{1}{2}\ln \left(1-\frac{1}{x^2}\right)$$
 [/INST] 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
$$\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{nx^{2n}}=-\log \left(1-\frac{1}{x^2}\right)$$
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