How to show that $\ln{x}=2\sum_{n=0}^{\infty}\frac{1}{(2n+1)}\left(\frac{x-1}{x+1}\right)^{2n+1}$?

Question

How do I show that $\displaystyle\ln{x}=2\sum_{n=0}^{\infty}\frac{1}{(2n+1)}\left(\frac{x-1}{x+1}\right)^{2n+1},\ x>0$?

I have tried using Mclaurin series of $\ln(1+x)$ and replacing $x$ with $\dfrac{x-1}{x+1}$ in it but that didn't worked.

Answer 1

Render

$\ln(1+y)=y-(y^2/2)+(y^3/3)-...$

$\ln(1-y)=-y-(y^2/2)-(y^3/3)-...$

Subtract, recognizing that the left side is $\ln[(1+y)/(1-y)]$. Thus put in

$x=(1+y)/(1-y)$

$\color{blue}{\ln(x)=2y+(2y^3/3)+...}$

and algebraically arrange this to give

$y=(x-1)/(x+1),$

which can be put into the right side of the blue equation above.

Answer 2

Simplify the expression inside the sum using an integral of a much more elementary function in the following way:

\begin{align*} 2\sum_{n=0}^{\infty}\frac 1{2n+1}\left(\frac{x-1}{x+1}\right)^{2n+1}&= 2\sum_{n=0}^{\infty}\int_0^{\frac{x-1}{x+1}}y^{2n}\,\mathrm dy= 2\int_0^{\frac{x-1}{x+1}}\sum_{n=0}^{\infty}y^{2n}\,\mathrm dy= 2\int_0^{\frac{x-1}{x+1}}\frac{\mathrm dy}{1-y^2}\\[10pt] &=\left.\ln\left(\frac{1+y}{1-y}\right)\right|_{y=0}^{y=\frac{x-1}{x+1}}=\ln x \end{align*}