How to calculate what this power series converges against? (double factorials) I'm working on my physics master course homework and I'm given the following equation out of nowhere:
$\displaystyle{
1 + \sum_{n\ =\ 1}^{\infty}{z^n\left(\, 2n - 1\,\right)!! \over 2n!!}
={1 \over \,\sqrt{\,\vphantom{\large A}1 - z\,}\,}
}$
Now I don't need to prove it for my Homework, but still im wondering,
how one would calculate this series.
Of course it is not mentioned in the Homework, that this series doesn't converge for
all $z$ ( sloppy physicsy style, I know : ) ). Wolfram Alpha said, It only converges for 
$\,{\rm abs}\left(\, z\,\right) < 1$, which is a hint to the geometric series, but I have no idea how to account for the double factiorals.
Any hints?

Derivation of the Series
The Binomial Theorem says that
$$
\begin{align}
(1-x)^{-1/2}
&=1+\frac{(-\frac12)}{1}(-x)^1+\frac{(-\frac12)(-\frac32)}{1\cdot2}(-x)^2+\frac{(-\frac12)(-\frac32)(-\frac52)}{1\cdot2\cdot3}(-x)^3+\dots\\
&=1+\frac12x+\frac{1\cdot3}{2\cdot4}x^2+\frac{1\cdot3\cdot5}{2\cdot4\cdot6}x^3+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8}x^4+\dots\\
&=1+\frac{1!!}{2!!}x+\frac{3!!}{4!!}x^2+\frac{5!!}{6!!}x^3+\frac{7!!}{8!!}x^4+\dots
\end{align}
$$
$n!!=n(n-2)(n-4)\dots(1\text{ or }2)$

Radius of Convergence
Note that
$$
1\le\frac{(2n+1)!!}{(2n)!!}=(2n+1)\frac{(2n-1)!!}{(2n)!!}\le(2n+1)
$$
therefore
$$
\frac1{2n+1}\le\frac{(2n-1)!!}{(2n)!!}\le1
$$
Using the formula for the Radius of Convergence, we get
$$
\frac1{\limsup\limits_{n\to\infty}1^{1/n}}\le r\le\frac1{\limsup\limits_{n\to\infty}\left(\frac1{2n+1}\right)^{1/n}}
$$
which gives a radius of convergence of $1$.
