How to derive the following series expansion for $\ln \left( \frac{x+1}{x-1} \right)$? 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 ?
 A: 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.
A: If $|x|>1$ then $\frac{1}{|x|}<1$, so:
$$\ln\left(\frac{x+1}{x-1}\right)=\ln\left(\frac{x(1+\frac{1}{x})}{x(1-\frac{1}{x})}\right)=\ln\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right)=\ln(1+t)-\ln(1-t)$$
With $t=\displaystyle\frac{1}{x}<1$, then $|t|<1$
Expanding :
$$\ln(1+t)=t-\frac{t^{^2}}{2}+\frac{t^{^3}}{3}-\frac{t^{^4}}{4}+...$$
$$\ln(1-t)=-t-\frac{t^{^2}}{2}-\frac{t^{^3}}{3}-\frac{t^{^4}}{4}-...$$
Then:
$$\ln(1+t)-\ln(1-t)=\left(t-\frac{t^{^2}}{2}+\frac{t^{^3}}{3}-\frac{t^{^4}}{4}+...\right)-\left(-t-\frac{t^{^2}}{2}-\frac{t^{^3}}{3}-\frac{t^{^4}}{4}-...\right)=$$
$$2\left(t+\frac{t^{^3}}{3}+\frac{t^{5}}{5}...\right)$$
Now $t=\displaystyle\frac{1}{x}$;
$$\ln\left(\frac{x+1}{x-1}\right)=2\left(\frac{1}{x}+\frac{1}{3x^{^3}}+\frac{1}{5x^{^5}}+\frac{1}{7x^{^7}}...\right)$$
