What is the limit of the following function?

Question

$\lim_{n\rightarrow \infty }\left ( n-1-2\left (\frac{\Gamma(n/2)}{\Gamma((n-1)/2)} \right )^2 \right )$

Answer 1

It is best to proceed using the following approximation for $\Gamma $ function $$\log \Gamma(x) =x\log x-x+\frac{1}{2}\log\frac{2\pi}{x}+\frac{1}{12(x+1)}+o(1/x)$$ and therefore we have $$2\log\frac{\Gamma(x/2)}{\Gamma ((x-1)/2)}=\log\frac{x-1}{2}+(x-1)\log\frac{x}{x-1}-1-\frac{1}{3(x+1)(x+2)}+o(1/x)$$ or $$2\log\frac{\Gamma (x/2)}{\Gamma ((x-1)/2)}=\log\frac{x-1}{2}-\frac{1}{2(x-1)}+o(1/x)$$ Exponentiating we get $$2\left(\frac{\Gamma (x/2)}{\Gamma ((x-1)/2)}\right)^{2}=(x-1)\exp\left(-\frac{1}{2(x-1)}+o(1/x)\right)$$ or $$2\left(\frac{\Gamma (x/2)}{\Gamma ((x-1)/2)}\right)^{2}=(x-1)-\frac{1}{2}+o(1)$$ It now follows that the desired limit $$\lim_{x\to\infty}\left((x-1)-2\left(\frac{\Gamma (x/2)}{\Gamma ((x-1)/2)}\right)^{2}\right)$$ is $1/2$. Note that it is important to deal with error terms properly to get a rigorous evaluation of the limit in question.

Answer 2

$$ \begin{aligned} \lim_{n\rightarrow \infty }\left (n-1-2\left (\frac{\Gamma(n/2)}{\Gamma((n-1)/2)} \right )^2 \right) & = \lim _{n\rightarrow \:\infty \:}\:\left(n-1-2\left(\frac{\frac{2\left(\frac{n}{2}\right)!}{n}}{\frac{2\left(\frac{n-1}{2}\right)!}{n-1}}\right)^2\right) \\& = \lim _{n\rightarrow \:\infty \:}\:\left(n-1-2\left(\frac{\frac{2\left(\sqrt{2\pi \:\left(\frac{n}{2}\right)}\left(\frac{\left(\frac{n}{2}\right)}{e}\right)^{\frac{n}{2}}\right)}{n}}{\frac{2\left(\sqrt{2\pi \:\:\left(\frac{n-1}{2}\right)}\left(\frac{\left(\frac{n-1}{2}\right)}{e}\right)^{\frac{n-1}{2}}\right)}{n-1}}\right)^2\right) \\& = \lim _{n\rightarrow \:\infty \:}\:\left(n-1-\frac{n^{n-1}\left(n-1\right)^{-n+2}}{e}\right) \\& = \lim _{t\rightarrow 0\:}\:\left(\frac{1}{t}-1-\frac{\left(\frac{1}{t}\right)^{\left(\frac{1}{t}\right)-1}\left(\frac{1}{t}-1\right)^{-\frac{1}{t}+2}}{e}\right) \\& = \lim _{t\rightarrow 0\:}\:\left(\frac{e-\left(-t+1\right)^{\frac{2t-1}{t}}}{et}-1\right) \\& = \lim _{t\rightarrow 0\:}\:\left(\frac{e-\left(e-\frac{3et}{2}+o\left(t\right)\right)}{et}-1\right) \\& = \color{red}{\frac{1}{2}} \end{aligned} $$

Solved with Stirling's approximation and Taylor expansion

Answer 3

Note that $$ {{\Gamma (n/2)} \over {\Gamma (n/2 - 1/2)}} = \left( {n/2 - 1/2} \right)^{\,\overline {\,1/2\,} } $$ where $x^{\,\overline {\,a\,}} $ denotes the Rising Factorial (rising Pochammer).

It is an increasing function for $1<n$.
Because of that and considering the rules for summing the exponents of the Rising Factorial, we have: $$ \left( {n/2 - 1/2} \right) = \left( {n/2 - 1/2} \right)^{\,\overline {\,1\,} } = \left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } \left( {n/2} \right)^{\,\overline {\,\,1/2\,\,} } > \left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2} $$ and $$ \left( {n/2 - 1} \right) = \left( {n/2 - 1} \right)^{\,\overline {\,1\,} } = \left( {n/2 - 1} \right)^{\,\overline {\,\,1/2\,\,} } \left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } < \left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2} $$ which means $$ \left( {n/2 - 1} \right) < \left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2} < \left( {n/2 - 1/2} \right) $$ and therefore we know that the given function is bound between $0$ and $1$ $$ 0 < n - 1 - 2\left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2} < 1\quad \left| {\;1 < n} \right. $$

The Stirling series for the rising factorial is $$ z^{\,\overline {\,w\,} } \propto z^{\,w} \left( {1 + {{w\left( {w - 1} \right)} \over {2\,z}} + O\left( {{1 \over {z^{\,2} }}} \right)} \right)\quad \left| \matrix{ \;\left| z \right| \to \infty \hfill \cr \;\left| {\arg (z + w)} \right| < \pi \hfill \cr} \right. $$ so we get $$ \eqalign{ & \left( {n/2 - 1/2} \right)^{\,\overline {\,1/2\,} } \propto \left( {n/2 - 1/2} \right)^{\,1/2\,} \left( {1 + {{\,1/2\,\left( {\,1/2\, - 1} \right)} \over {2\,\left( {n/2 - 1/2} \right)}} + O\left( {{1 \over {n^{\,2} }}} \right)} \right) = \cr & = \left( {{{n - 1} \over 2}} \right)^{\,1/2\,} \left( {1 - {{\,1} \over {4\,\left( {n - 1} \right)}} + O\left( {{1 \over {n^{\,2} }}} \right)} \right) \cr} $$ and finally $$ \eqalign{ & n + 1 - 2\left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,1/2\,} } } \right)^{\,2} \propto n + 1 - \left( {n - 1} \right)\left( {1 - {{\,1} \over {2\,\left( {n - 1} \right)}} + O\left( {{1 \over {n^{\,2} }}} \right)} \right) = \cr & = {1 \over 2} + O\left( {{1 \over n}} \right) \cr} $$