Show that a function is equal to this series Show that $$ (1-x)^{-1/2} = \sum_{n=0}^{\infty} \frac{{(2n)}!}{{{(n}!)}^2} \ \left ( \frac{x}{2} \right )^{2n}  $$
I have tried u substitution: $ u = 1-x^2 $. When I calculated $f(0)$, I got an undefined term. I have tried building off common Macluarin series such as $ \frac{1}{1-x}$ with no luck. 
 A: The $n$th derivative is 
$$
 ((1-x)^{-1/2})^{(n)}=\frac{(2n-1)!!}{2^n}\frac1{(1-x)^{(2n+1)/2}}
$$
So the Macluarin series is
$$
(1-x)^{-1/2}=\sum_{n=0}^{\infty} \frac{(2n-1)!!}{2^n}\frac{x^n}{n!}=\sum_{n=0}^{\infty} \frac{(2n)!}{2^n2\cdots2n}\frac{x^n}{n!} =\sum_{n=0}^{\infty} \frac{(2n)!}{(n!)^2}\left(\frac{x}{2^2}\right)^n
$$
A: Using Binomial series,
$$(1-x)^{-1/2}=\sum_{r=0}^\infty\dfrac{-1/2(-1/2-1)\cdots\{-1/2-(r-1)\}}{r!}(-x)^r$$
Now $\displaystyle-1/2(-1/2-1)\cdots\{-1/2-(r-1)\}$
$\displaystyle=\left(-\dfrac12\right)^r1\cdot3\cdot5\cdots(2r-1)$
$\displaystyle=\left(-\dfrac12\right)^r\dfrac{(2r)!}{2^r r!}$
A: I assume there is a typo in your question. Another way is to note that the Taylor series of the arcsin is $$ \arcsin\left(x\right)=\sum_{n\geq0}\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^{2}\left(2n+1\right)}x^{2n+1}
 $$ hence if we derivate $$\left(1-x^{2}\right)^{-1/2}=\sum_{n\geq0}\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^{2}}x^{2n}=\sum_{n\geq0}\frac{\left(2n\right)!}{\left(n!\right)^{2}}\left(\frac{x}{2}\right)^{2n}.$$
