Prove power series of $\ln\left(\frac{2 - 2\sqrt{1-x}}{x}\right)$

Question

$$\ln\left(\frac{2(1 - \sqrt{1-x})}{x}\right) = \frac{1}{2}\frac{x}{2} + \frac{1}{2}\frac{3}{4}\frac{x^2}{4} + \frac{1}{2}\frac{3}{4}\frac{5}{6}\frac{x^3}{6} + \cdots\ \ \ \ \text{where} |x| < 2$$

I transformed $\ln\left(\frac{2(1 - \sqrt{1-x})}{x}\right)$ to $\ln(2) + \ln(1 - \sqrt{1-x}) + \ln(x)$ and got stuck.

I know that have to calculate Taylor series of $\ln(1 - \sqrt{1-x})$ but I do not have idea how to do that.

Answer 1

Hint

May be, you could start using the generalized binomial theorem or Taylor series to expand $$\sqrt{1-x}=1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}+O\left(x^4\right)$$ which makes $$1-\sqrt{1-x}=\frac{x}{2}+\frac{x^2}{8}+\frac{x^3}{16}+O\left(x^4\right)$$ and continue until you arrive to something looking like $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$

Answer 2

A tricky approach is to start from the RHS, namely $$ \sum_{n\geq 1}\frac{(2n-1)!!}{(2n)!!}\cdot\frac{x^n}{2n} = \sum_{n\geq 1}\frac{(2n)!}{4^n n!^2}\cdot\frac{x^n}{2n}=\frac{1}{2}\int_{0}^{x}\sum_{n\geq 1}\frac{(2n)!}{4^n n!}t^{n-1}\,dt $$ then recognize, through the extended binomial theorem, $$ \frac{1}{2}\int_{0}^{x}\frac{1-\sqrt{1-t}}{t\sqrt{1-t}}\,dt =\log\left(\frac{2}{1+\sqrt{1+x}}\right).$$