Pochhammer polynomials

Series expansion of $\displaystyle \frac{1}{\sqrt{1-x}}$

We prove the following result posted by @infseriesbot
\[ \frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!} x^n\]

Proof

\[ \frac{1}{\sqrt{1-x}} = (1-x)^{-\frac{1}{2}} \] Recall that Newton extended the binomial theorem to allow real and complex numbers as exponents: \[(x+y)^r = \sum_{k=0}^{\infty} \binom{r}{k}x^{r-k}y^k \quad |x|>|y|\] where \[\binom{r}{k} = \frac{(r)_{k}}{k!} \] and $\displaystyle (r)_{k}$ is the Pochhammer symbol (falling factorial): \[(r)_{k} = r(r-1)\cdots(r-k+1) \] Hence: \begin{align*} (1-x)^{-\frac{1}{2}} =& \sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n} (-1)^n x^{n} \\ =& \sum_{n=0}^{\infty} \frac{\left(-\frac{1}{2}\right)_{n}}{n!} (-1)^n x^{n} \\ \end{align*} One of the properties of the rising and falling factorial states that \[ r^{(n)} = (-1)^n(r)_{n}\] where $\displaystyle r^{(n)}$ is the rising factorial: \[r^{(n)} = r(r+1)(r+2)\cdots (x+n-1)\] Hence \[ \sum_{n=0}^{\infty} \frac{\left(-\frac{1}{2}\right)_{n}}{n!} (-1)^n x^{n} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}}{n!} x^{n}\] Note that we can express the factorial as a the pochhammer polynomial of $(1)^{(n)}$: \[(1)^{(n)} = 1\cdot 2\cdot 3\cdots(n-1)n = n!\] Hence, \[ \frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}}{n!} x^{n} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}}{(1)^{(n)}} x^{n} \] On the other hand, double factorials can be expressed with the following form: \[(2n-1)!! = 2^n \frac{\Gamma\left(\frac{1}{2}+n\right)}{\sqrt{\pi}} = 2^n \frac{\Gamma\left(\frac{1}{2}+n\right)}{\Gamma\left(\frac{1}{2}\right)} \] \[(2n)!! = 2^n n! = 2^n \Gamma(n+1) = 2^n \frac{\Gamma(n+1)}{\Gamma(1)}\] while the pochhammer polynomial can be expressed as the quotient of two gamma functions: \[ r^{(n)} = \frac{\Gamma(n+r)}{\Gamma(r)}\] Hence, we can write: \[(2n-1)!!= 2^n \left(\frac{1}{2}\right)^{(n)}\] \[(2n)!! = 2^n (1)^{(n)}\] Therefore \[\frac{(2n-1)!!}{(2n)!!} = \frac{2^n \left(\frac{1}{2}\right)^{(n)}}{2^n (1)^{(n)}} = \frac{\left(\frac{1}{2}\right)^{(n)}}{(1)^{(n)}}\] Hence, we can conclude \[ \frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}}{(1)^{(n)}} x^{n} = \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!} x^{n}\] This can also be expressed with the notation of the generalized hypergeometric function as \[\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}}{(1)^{(n)}} x^{n} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}(1)^{(n)}}{(1)^{(n)}} \frac{x^{n}}{n!} = {}_{2}F_{1}\left(\frac{1}{2},1;1;x\right)\] However, in the hypergeometric function notation it is common to use $(x)_{n}$ to denote the rising factorial instead of the falling factorial, so it is very important to always check if we are referring to rising of falling factorials. \[\boxed{\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!} x^{n}} \]