Asymptotic behaviour of $1- \left( \frac{\Gamma(n+\frac{1}{2})}{\sqrt{n} \Gamma(n)} \right) ^2$

Question

I know that $$\lim_{n\rightarrow \infty}\frac{\Gamma(n+\frac{1}{2})}{\sqrt{n} \Gamma(n)}=1,$$ but I'm interested in the exact behaviour of

$$a_n =1- \left( \frac{\Gamma(n+\frac{1}{2})}{\sqrt{n} \Gamma(n)} \right) ^2$$

particularily compared to $$b_n = \frac{1}{4n}$$

I haven't studied asymptotics yet, so I have no idea how to approach this, but I need this particular result in a statistics problem I'm working on.

Answer 1

The following code in Mathematica

Series[1 - (Gamma[x + 1/2]/(Sqrt[x] Gamma[x]))^2, {x, Infinity, 6}]

gives $$ \frac{1}{4 x}-\frac{1}{32 x^2}-\frac{1}{128 x^3}+\frac{5}{2048 x^4}+\frac{23}{8192 x^5}-\frac{53}{65536 x^6}+O\left[\frac{1}{x}\right]^7 $$

Answer 2

Using Stirling formula you can show that $$\frac{\Gamma(x+\frac12)}{\sqrt{x}\,\Gamma(x)}=1-\frac{1}{8x}+\frac{1}{128x^2}+O(x^{-3})$$ as $x\rightarrow \infty$. In principle one can obtain as many terms in the asymptotics as one wants using the formula with Bernoulli numbers in the above link.

Answer 3

Hint: Use Stirling approximation

$$ n!=\Gamma(n+1) \sim \left(\frac{n}{e}\right)^n\sqrt{2 \pi n}. $$

Added: If you use the above approximation, you will get

$$ a_n\sim 1-{\frac {{{\rm e}^{-1}} \left( n-1 \right) ^{1-2\,n}{4}^{-n} \left( - 1+2\,n \right) ^{2\,n}}{n}} .$$

Double check the the calculations.