Limiting value of the coefficients of the power series $(1-x)^{-\frac{1}{2}}$ Suppose we have the power series $(1-x)^{-\frac{1}{2}}$. We know that this power series converges iff $|x|<1$. Suppose $a_n$ denote the coefficieis nt of this power series. So $a_0=1, a_1=\frac{1}{2}, a_2= \frac{3}{8}, a_3=\frac{5}{16} \ldots \text{ so on}$. My question is what is $\lim_{n\to \infty} a_n$. It is clear that $a_n>0$ for every $n$ and $a_n$ is a decreasing sequence, so limit does exist. But I couldn't find the exact value of the limit. Since for $x=1$ the power series doesn't converge, we cannot say anything from there.
Thanks in advance for any kind of help.
 A: Determine a general formula for the coefficients
How do we compute these coefficients? The way to expand this is: 
$$(1-x)^{-1/2}=\frac1{0!}(-x)^0+\frac1{1!}(-x)^1\left(-\frac12\right)+\frac1{2!}(-x)^2\left(-\frac12\right)\left(-\frac32\right)+\frac1{3!}(-x)^3\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)+\cdots$$
Notice that as we go from $a_n$ to $a_{n+1}$, we multiply  each coefficient by $-\frac1{n+1}\left(-\frac12-n\right)=\frac1{n+1}\left(\frac{2n+1}2\right)$. So We have a recurrance relation $$\begin{align}a_{n+1}&=\frac{2n+1}{2(n+1)}a_n\\&=\frac{(2n+1)(2n-1)\cdots(1)}{2^{n+1}(n+1)!}a_0\\&=\frac{(2n+1)!}{2^{n+1}(n+1)!}\cdot\frac{1}{(2n)(2n-2)\cdots(2)}\\&=\frac{(2n+1)!}{2^{n+1}(n+1)!}\cdot\frac{1}{2^nn!}\\&=\frac{(2n+1)!}{2^{2n+1}n!(n+1)!}\end{align}$$
So we have, for $n>0$ $$a_n=\frac{(2n-1)!}{2^{2n-1}n!(n-1)!}$$

Evaluate coefficients in large $n$ limit
Stirling's approximation tells us that for large $n$, $\,\,n!\sim n^{1/2}n^ne^{-n}$. Using this, for large $n$, $$\begin{align}a_n&\sim\left(\frac{2n-1}{n(n-1)}\right)^{1/2}\frac{2^{2n-1}n^{2n-1}e^{-2n+1}}{2^{2n-1}n^ne^{-n}n^{n-1}e^{-n+1}}\\&\sim\left(\frac{2n-1}{n^2-n}\right)^{1/2}\\&\sim\left(\frac2n\right)^{1/2}\to 0\end{align}$$
So $$\lim_{n\to\infty}a_n=0$$
A: Notice the binomial coefficents $${p\choose 0}=1, {p\choose 1}= p,~ {p\choose 2}=\frac{p(p-1)}{2},~ {p\choose 3}= \frac{p(p-1)(p-2)}{3!}$$ $$,~ {p \choose 4}=\frac{p(p-1)(p-2)(p-3)}{4!}~~~(1)$$
where $p$ may be positive/negative integer or a fraction.
Binomial infinte series series when $|z|<1$ is given as
$$(1+z)^{p}= \sum_{k=0}^{\infty} {p \choose k} z^k$$ 
Using  $p=-1/2$, $z=-x$ and calculating the binomial coefficents as in (1), we get
$$(1-x)^{-1/2}=1-\frac{-1}{2} (-x)+ \frac{(-1/2)(-1/2-1)}{2!} (x)^2+ \frac{(-1/2)(-1/2-1)(-1/2-2)}{3!}+...ad-inf.$$
$$\implies (1-x)^{-1/3}=1+\frac{x}{3}+\frac{3}{8}x^2+\frac{5}{16}x^3+ \frac{35}{128}x^4+.....+..ad-inf $$
A: $$
a_{\,n}  = \left( { - 1} \right)^{\,n} \left( \matrix{
   - 1/2 \cr 
  n \cr}  \right) = \left( \matrix{
  n - 1/2 \cr 
  n \cr}  \right) = {{\Gamma \left( {n + 1/2} \right)} \over {\Gamma \left( {n + 1} \right)\Gamma \left( {1/2} \right)}}
$$
and
$$
\mathop {\lim }\limits_{n\, \to \;\infty } a_{\,n}  = {1 \over {\Gamma \left( {1/2} \right)}}\mathop {\lim }\limits_{n\, \to \;\infty }
 {1 \over {{{\Gamma \left( {n + 1/2 + 1/2} \right)} \over {\Gamma \left( {n + 1/2} \right)}}}} \approx {1 \over {\sqrt \pi  }}{1 \over {\sqrt n }} = 0
$$
