What is the limitation of $\Gamma(\frac{n+1}{2})/\Gamma(\frac{n}{2})$?

Question

What is the limitation of $$\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}$$ as $n \rightarrow \infty$?

If it diverges, could I get its divergence rate?

$\Gamma(x)$ is the gamma function here.

Answer 1

Stirling's formula

$$\log \Gamma(z) = \bigl(z-\tfrac{1}{2}\bigr) \log z - z + \tfrac{1}{2} \log (2\pi) + O\left(\frac{1}{\operatorname{Re} z}\right)$$

gives us

\begin{align} \log \Gamma\bigl(z+\tfrac{1}{2}\bigr) - \log \Gamma(z) &= z\log\bigl(z+\tfrac{1}{2}\bigr) - \bigl(z-\tfrac{1}{2}\bigr)\log z - \tfrac{1}{2} + O\left(\frac{1}{\operatorname{Re} z}\right)\\ &= \tfrac{1}{2}\log z - \tfrac{1}{2} + z\log\left(1 + \tfrac{1}{2z}\right) + O\left(\frac{1}{\operatorname{Re} z}\right)\\ &= \tfrac{1}{2}\log z + O\left(\frac{1}{\operatorname{Re} z}\right) \end{align}

and hence

$$\frac{\Gamma\bigl(\frac{n+1}{2}\bigr)}{\Gamma\bigl(\frac{n}{2}\bigr)} \sim \sqrt{\frac{n}{2}}.$$

Answer 2

Let: $$A_n\triangleq\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}.$$ Then: $$ A_n A_{n+1}=\frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(\frac{n+2}{2}\right)}{\Gamma\left(\frac{n}{2}\right)\Gamma\left(\frac{n+1}{2}\right)}=\frac{n}{2},$$ so $$ A_n\sim\sqrt\frac{n}{2},$$ since $\log\Gamma$ is convex.