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Assume $n \in \mathbb{N}$. Using Wolfram Alpha I found that the following has a nice limit: $$ \begin{align*} \lim_{n \rightarrow \infty} \frac{2\sqrt{2}\cdot\Gamma(\frac{n+3}{2})}{n^{3/2}\cdot\Gamma(\frac{n}{2})} =1 \end{align*} $$ However, I'm not really sure how to arrive to that result through half-integer gamma.

From what I'm gathering,

$$ \frac{\Gamma(\frac{n+3}{2})}{\Gamma(n/2)} = 2^{-3/2}\frac{(n+1)!!}{(n-2)!!} $$ which, I can only guess, should evaluate at $2^{-3/2} n^{3/2}$ or something close, but cannot see why.

RobPratt
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runr
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1 Answers1

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If you're open to other methods, using Stirling's approximation,

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

you have

$$\operatorname{\Gamma}\left(\frac{n+3}2\right)\sim\sqrt{\pi(n+3)}\left(\frac{n+3}{2e}\right)^{\frac{n+3}2}\\ \operatorname{\Gamma}\left(\frac n2\right)\sim\sqrt{\pi n}\left(\frac n{2e}\right)^{\frac n2}$$

and in the limit,

$$\begin{align} \lim_{n\to\infty}\frac{2\sqrt2\operatorname{\Gamma}\left(\frac{n+3}2\right)}{n^{\frac32}\operatorname{\Gamma}\left(\frac n2\right)}&=\lim_{n\to\infty}\frac{2\sqrt2\sqrt{\pi(n+3)}\left(\frac{n+3}{2e}\right)^{\frac{n+3}2}}{n^{\frac32}\sqrt{\pi n}\left(\frac n{2e}\right)^{\frac n2}}\\[1ex] &=\frac{2\sqrt2}{(2e)^{\frac32}}\lim_{n\to\infty}\left(\frac{n+3}n\right)^{\frac{n+3}2}\\[1ex] \end{align}$$

and $\left(\frac{n+3}n\right)^{\frac{n+3}2}\to e^{\frac32}$, so the limit of $1$ follows.

user170231
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  • Both yours and @metamorphy linked one are very interesting approaches, learned new tricks today, thanks! – runr Feb 05 '21 at 22:49