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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 $|x|<1$?2012-09-20
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    Other useful facts for this problem: $1 - 1/x^2 = (1 - 1/x)(1 + 1/x)$ and $\ln(ab) = \ln(a) + \ln(b)$.2012-09-20

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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

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

  • 0
    To prove Taylor's series, do we just check that both sides agree at zero, and their derivatives agree everywhere?2012-09-21
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    @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).2012-09-21