How to calculate what this power series converges against? (double factorials)

Question

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?

Answer 1

Derivation of the Series

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

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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 $1$.

Answer 2

Here is one way you can attack this problem. I will not write the full analysis, but enough to give an idea. Start with assuming we have $$ \frac1{\sqrt{1-z}} = \sum_{k=0}^\infty a_k z^k $$ Then, computing the square, we find that $$ \frac1{\sqrt{1-z}} \cdot \frac1{\sqrt{1-z}} = \frac1{1-z}= \sum_{k=0}^\infty a_k z^k \cdot \sum_{l=0}^\infty a_l z^l $$ But that must be equal the usual geometric series, so the problem can be seen as "computing the (convolutional) square root of a power series. We get that $$ \sum_{k=0}^\infty \sum_{l=0}^k a_k a_{l-k} =1, $$ which then can be solved recursively.