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This is a continuation of another question I asked. It seems only incidentally related to statistics, so I figured it would be better separate. (Also, I'm not able to make nearly as much progress on this one.)

Let

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

What is the limit of $c$ as $n \rightarrow \infty$?

My TI-83 blows up around $n=100$, and convergence there seems pretty slow, so I can't say for sure. But it looks as though the answer could be $1$.

As for showing that... I've tried breaking it up into two cases: one where $n$ is odd and one where it's even. More accurately, I got something for odd $n$ and didn't see much point in exploring the even case, since it would probably look just as inscrutable.

If $n$ is the $k$th odd number — i.e., $n = 2k - 1$, then $\Gamma\left(\frac{n-1}{2}\right)=\Gamma(k-1)=(k-2)!$ and

$$\Gamma(n/2)=\left[\prod_{i=1}^{k-1} (2i-1)\right]\frac{\sqrt{\pi}}{2^{k-1}}$$

This can be rewritten as

$$\frac{\sqrt{\pi}}{2^{k-1}}\cdot\frac{[2(k-1)]!}{(k-1)!2^{k-1}}$$

The ratio of the two gamma functions, then, can be written as

$$\frac{[(k-1)!]^2 2^{2k-2}}{\sqrt{\pi}[2(k-1)]!(k-2)}$$

Without getting into specifics, I could kinda see a lot of the terms on top being absorbed by those in the bottom, but then I guess if I'm saying it converges to $1$, I wouldn't want them to be absorbed too much.

If anyone can offer ideas for the next step, I'd very much appreciate it.

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dmk
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    Stirling's formula should get you to an answer relatively painlessly. – Antonio Vargas Jan 29 '14 at 04:07
  • @Antonio Vargas: Thanks for the suggestion. I looked up the formula and figured out an example problem. However, for this one, I get $\sqrt{e}$. I used $\Gamma(n)=\sqrt{2\pi/n}(n/e)^n$, plugging in $k-1$ and $k$. (I can leave out $\mathcal{O}(1/n)$, right?) Since I don't have room here, I'll just say that since $e$ appears both in the numerator and in the denominator, each time raised to a different power, I don't see how to get it to cancel out. Wolfram confirms that the answer is $1$ but tells me the step-by-step solution is "unavailable". What should I do? – dmk Jan 29 '14 at 21:19
  • I figured out where my error was: Simplifying yields $\lim_{n\rightarrow\infty}\sqrt{e}\left(\frac{n-1}{n}\right)^{(n-1)/2}$; I had been thinking, Well, that $(n-1)/n$ thing looks like it goes to $1$. Rewriting it as $1-\frac{1}{n}$ made it a lot more obvious that it converges to some power of $e$ -- namely, $\sqrt{1/e}$. Therefore, everything cancels out, and the desired result, $1$, follows. – dmk Jan 31 '14 at 21:07

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The $\Gamma$ function is a log-convex function as a consequence of its definition through the Bohr-Mollerup theorem or the integral version of the Cauchy-Schwarz inequality: $$ \Gamma\left(\frac{\alpha+\beta}{2}\right)^2 = \left(\int_{0}^{+\infty}x^{\alpha+\beta}\frac{e^{-x}\,dx}{x}\right)^2\leq \Gamma(\alpha)\,\Gamma(\beta). $$

A nice consequence of the log-convexity is Gautschi's inequality: $$ x^{1-s} \leq \frac{\Gamma(x+1)}{\Gamma(x+s)}\leq (1+x)^{1-s} $$ that easily proves that your limit is $\color{red}{\large 1}$ by squeezing.

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Jack D'Aurizio
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