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How to prove that :

$$ \sum_{n=1}^{\infty}\frac{1}{2nx^{2n}}=-\frac{1}{2}\ln \left(1-\frac{1}{x^2}\right)$$

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Do you know that $\sum \frac1nx^n = -\ln(1-x)$ for $|x|<1$? – Hagen von Eitzen Sep 20 '12 at 20:41
Other useful facts for this problem: $1 - 1/x^2 = (1 - 1/x)(1 + 1/x)$ and $\ln(ab) = \ln(a) + \ln(b)$. – Michael Joyce Sep 20 '12 at 20:49

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 $-1\le x < 1$

$$-\log (1-x)=\sum^{\infty}_{n=1} \frac{x^n}n$$


$$-\log \left(1-\frac{1}{x}\right)=\sum^{\infty}_{n=1} \frac{1}{n x^n}$$

$$-\log \left(1-\frac{1}{x^2}\right)=\sum^{\infty}_{n=1} \frac{1}{n x^{2n}}$$

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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To prove Taylor's series, do we just check that both sides agree at zero, and their derivatives agree everywhere? – Frank Sep 21 '12 at 7:44
@MohammedAl-mubark Using the geometric series $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$ for $|x|<1$, we integrate the sum term by term and the result follows (plug in a known value of $x$ to both sides to find the constant of integration). – Argon Sep 21 '12 at 21:01

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