McLaurin expansion of $(a^2+z^2)^{-\frac{1}{2}}$

Question

Show that $$(a^2+z^2)^{-\frac{1}{2}}=\frac{1}{a}+\sum_\limits{n=1}^{\infty}\frac{(-1)^n(2n-1)!!)}{a^{2n+1}(2n)!!}z^{2n},\:|z|<|a|,a>0$$

$$(a^2+z^2)^{-\frac{1}{2}}=\frac{1}{a}(1+\frac{z^2}{a^2})^{-\frac{1}{2}}=\frac{1}{a}(1+\frac{z^2}{a^2})^{-\frac{1}{2}}=\frac{1}{a}+\frac{1}{a}\sum_\limits{n=1}^{\infty}{-\frac{1}{2}\choose n}(\frac{z^2}{a^2})^n$$

Now my problem lies in the simplification of ${-\frac{1}{2}\choose n}$

$${-\frac{1}{2}\choose n}=\frac{-\frac{1}{2}!}{n!(-n-\frac{1}{2})!}=\frac{(-1)^n(\frac{1}{2}(\frac{1}{2}+1)...(\frac{1}{2}+n-1))}{n!}$$

Now I am stuck and I do not know how to simplify the expression any longer.

Question:

Can someone help me solve the exercise?

Thanks in advance!

Answer 1

Hint. This might be useful \begin{align} \frac{(2n-1)!!}{(2n)!!} &=\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n}\\\\ &=\frac{1\cdot \color{red}{2}\cdot 3\cdot \color{red}{4} \cdot 5\cdot\color{red}{6}\cdots(2n-1)\cdot \color{red}{2n}}{(2\cdot 4\cdot 6\cdots (2n))^\color{red}{2}}\\\\ &=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^\color{red}{2}}\\\\ & =\frac{(2n)!}{2^{2n} (n!)^2 }. \end{align}