How to show that $\sum_{n=0}^\infty\frac{(2n)!}{(n+1)!n!}(\frac{a}{2})^{2n}\sum_{m=0}^n b^{2m}=\frac{1}{\sqrt{1-a^2}+\sqrt{1-a^2b^2}}$? How to show that $$\sum_{n=0}^\infty\frac{(2n)!}{(n+1)!n!}(\frac{a}{2})^{2n}\sum_{m=0}^n b^{2m}=\frac{1}{\sqrt{1-a^2}+\sqrt{1-a^2b^2}}$$
I substitute $\sum_{m=0}^n b^{2m}=\frac{1-b^{2n+2}}{1-b^2}$ into the equation. I tried really hard and I have no clue about what to do to derive the square root.
Could anyone kindly help? Thanks!
 A: The right-hand side of the equation should actually be 
$$\frac{\color{red}{2}}{\sqrt{1 - a^2} + \sqrt{1 - a^2b^2}}.\tag{*}$$
I'll assume $a$ and $b$ are real numbers such that $|a| < 1$ and $|ab| < 1$. Since 
$$(2n)! = 2^{2n}n!\left(n - \frac{1}{2}\right)\cdots \left(\frac{3}{2}\right)\left(\frac{1}{2}\right)$$
and
$$\left(n - \frac{1}{2}\right)\cdots \left(\frac{3}{2}\right)\left(\frac{1}{2}\right) = (-1)^n\left(\frac{1}{2} - n\right)\cdots \left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right) = n!\binom{-1/2}{n},$$
then 
$$\sum_{n = 0}^\infty \frac{(2n)!}{(n+1)!n!}\left(\frac{a}{2}\right)^{2n}\sum_{m = 0}^n b^{2m} = \sum_{n = 0}^\infty \frac{(-1)^n}{n+1}\binom{-1/2}{n}a^{2n}\frac{1 - b^{2n+2}}{1-b^2}.\tag{**}$$
Let $F(a,b)$ represent the series in (**). If $a = 0$, then $F(a,b) = 1$, which agrees with formula (*) when $a = 0$. If $a\neq 0$, then 
$$\begin{align}F(a,b) &= \frac{1}{a^2(1 - b^2)}\sum_{n = 0}^\infty \frac{(-1)^n}{n+1}\binom{-1/2}{n}[(a^2)^{n+1} - (a^2b^2)^{n+1}]\\
&= \frac{1}{a^2(1-b^2)}\sum_{n = 0}^\infty (-1)^n\binom{-1/2}{n}\int_{a^2b^2}^{a^2} x^n\, dx\\
&= \frac{1}{a^2(1-b^2)}\int_{a^2b^2}^{a^2}\sum_{n = 0}^\infty \binom{-1/2}{n}(-x)^n\, dx\\
&= \frac{1}{a^2(1-b^2)}\int_{a^2b^2}^{a^2} (1 - x)^{-1/2}\, dx\\
&= \frac{1}{a^2(1-b^2)}[2\sqrt{1-a^2b^2} - 2\sqrt{1 - a^2}]\\
&= \frac{2}{a^2(1-b^2)}\frac{a^2 - a^2b^2}{\sqrt{1-a^2}+\sqrt{1-a^2b^2}}\\
&= \frac{2}{\sqrt{1-a^2}+\sqrt{1-a^2b^2}}.
\end{align}$$
