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How to prove :

$$ \ln \biggl(\frac{x+1}{x-1}\biggr) = 2\biggl[\frac{1}{x} + \frac{1}{3x^3} + \frac{1}{5x^5} + \cdots \biggr]$$ where $|x| \gt 1$

I am not able to get how to proceed on this one ?

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Typo alert: you probably mean $5x^5$, no? – Alex Basson Nov 21 '10 at 14:22
Yes,I have fixed it now. :) – Quixotic Nov 21 '10 at 14:25
What happens if you replace $x$ in $\frac{1+x}{1-x}$ with $\frac1{x}$? – J. M. Nov 21 '10 at 14:30
@J.M: Superb! I got the desired result by substituting $x$ by $1/x$ in the expansion of $\log \left(\frac{1+x}{1-x}\right) = 2\biggl[\frac{x}{1} + \frac{x^3}{3} + \frac{x^5}{5} + \cdots \biggr] $ – Quixotic Nov 21 '10 at 14:36
@Sri: This was algebra-precalculus. – Quixotic Dec 13 '11 at 9:59
up vote 3 down vote accepted

If $|x|>1$ then $\frac{1}{|x|}<1$, so:


With $t=\displaystyle\frac{1}{x}<1$, then $|t|<1$

Expanding :






Now $t=\displaystyle\frac{1}{x}$;


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Note that $\frac{x+1}{x-1} = \frac{1 + 1/x }{1 - 1/x}$.

Note that the series for $\log (1+x)$ converges absolutely when $|x| <1$. Therefore it is permissible to rearrange the series in any way you like.

Now note that $\log \left(\frac{1+1/x}{1-1/x}\right) = \log (1+1/x) - \log (1-1/x)$. Expand both logarithmic series and group the terms.

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$\log \left(\frac{1+x}{1-x}\right) = \log (1+x) - \log (1-x) = 2\biggl[\frac{x}{1} + \frac{x^3}{3} + \frac{x^5}{5} + \cdots \biggr]$ – Quixotic Nov 21 '10 at 14:34

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