Evaluating $\lim_{n \to \infty}\left[\,\sqrt{\,2\,}\,\frac{\Gamma\left(n/2 + 1/2\right)}{\Gamma\left(n/2\right)} - \,\sqrt{\,n\,}\right]$

Question

I am trying to find this limit: $$\lim_{n \to \infty}\left[\,\sqrt{\,2\,}\,\frac{\Gamma\left(n/2 + 1/2\right)}{\Gamma\left(n/2\right)} - \,\sqrt{\,n\,}\right]$$

I know that $$\frac{\sqrt{2}\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})}=\begin{cases}\sqrt{\frac{\pi}{2}} \frac{(n-1)!!}{(n-2)!!},\ n\ is\ even\sim \sqrt{n-1}\\ \sqrt{\frac{2}{\pi}} \frac{(n-1)!!}{(n-2)!!},\ n\ is\ odd \sim\sqrt{n+1}\end{cases}$$ However, I can not find the limit above. Based on the numeric study, it seems like the limit is 0:

Answer 1

$$f(n)=\sqrt{2}\,\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})} - \sqrt{n}=\sqrt{2}\,y- \sqrt{n}$$

Take logarithms $$\log(y)=\log \left(\frac{\Gamma \left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n}{2}\right)}\right)=\log \left(\Gamma \left(\frac{n+1}{2}\right)\right)-\log \left(\Gamma \left(\frac{n}{2}\right)\right)$$ Use Stirling approximation twice $$\log(y)=\frac{1}{2} \log \left(\frac{n}{2}\right)-\frac{1}{4 n}+\frac{1}{24 n^3}+O\left(\frac{1}{n^5}\right)$$ Continue with Taylor $$y=e^{\log(y)}=\sqrt{\frac n 2}\Bigg[1-\frac{1}{4 n}+\frac{1}{32 n^2}+O\left(\frac{1}{n^3}\right)\Bigg]$$ $$f(n)=-\frac 1{4\sqrt {n}}\Bigg[1-\frac{1}{8 n}-\frac{5}{32 n^2}+O\left(\frac{1}{n^3}\right)\Bigg]$$

Use it for $n=9$; rhe above truncated series gives $$f(9) \sim -\frac{2551}{31104}=-0.0820152\cdots$$ while the exact value is $$f(9)=-3+\frac {128}{35} \sqrt{\frac 2 \pi}=-0.0820222\cdots$$

Answer 2

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\begin{align} &\bbox[5px,#ffd]{\lim_{n \to \infty}\bracks{\root{2}{\Gamma\pars{n/2 + 1/2} \over \Gamma\pars{n/2}} - \root{n}}} = \lim_{n \to \infty}\bracks{\root{2}{\pars{n/2 - 1/2}! \over \pars{n/2 - 1}!} - \root{n}} \\[5mm] = & \ \lim_{n \to \infty}\bracks{\root{2}{\root{2\pi}\pars{n/2 - 1/2}^{n/2}\expo{-n/2 + 1/2} \over \root{2\pi}\pars{n/2 - 1}^{n/2 - 1/2}\,\expo{-n/2+1}} - \root{n}} \ \substack{Stirling\ Asymptotic\\[1mm] Behavior} \\[5mm] = & \ \lim_{n \to \infty}\bracks{\root{2}{\pars{n/2}^{n/2}\,\pars{1 - 1/n}^{n/2} \over \pars{n/2}^{n/2 - 1/2}\,\,\,\pars{1 - 2/n}^{n/2}}\expo{-1/2} - \root{n}} \\[5mm] = & \ \lim_{n \to \infty}\bracks{\root{2}{\expo{-1/2} \over \root{2/n}\expo{-1}}\expo{-1/2} - \root{n}} = {\large 0} \end{align}

Answer 3

Using the first few terms of http://dlmf.nist.gov/5.11.E13 with $a=1/2$, $b=0$ and $z=n/2$, we find $$ \frac{{\Gamma (n/2 + 1/2)}}{{\Gamma (n/2)}} = \sqrt {\frac{n}{2}} \left( {1 - \frac{1}{{4n}} + \mathcal{O}\!\left( {\frac{1}{{n^{2 } }}} \right)} \right). $$ Thus, $$ \sqrt 2 \frac{{\Gamma (n/2 + 1/2)}}{{\Gamma (n/2)}} - \sqrt n = - \frac{1}{{4\sqrt n }} + \mathcal{O}\!\left( {\frac{1}{{n^{3/2} }}} \right) $$ and your limit is indeed $0$.