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I am trying to write the Wallis Product (WP) using the $\Gamma$ function, thereby hoping for either a new identity or a simple derivation of the formula, but I get a result I am not sure its correct.

This is the WP
$$ \frac{\pi}{2} = \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}\cdots\frac{2n}{2n-1}\frac{2n}{2n+1}\cdots $$

Now $$ \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}\cdots\frac{2n}{2n-1}\frac{2n}{2n+1}\cdots = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdot \dots\ (2n)(2n)}{1\cdot1\cdot3\cdot3\cdot5\cdot5\cdot\dots\ (2n-1)(2n-1)(2n+1)}\cdots $$

$$=\frac{1\cdot2\cdot1\cdot2\cdot2\cdot2\cdot2\cdot2\cdot3\cdot2\cdot3\cdot2\cdot \hspace{12pt} \dots\hspace{12pt}n\cdot2\hspace{12pt}\cdot\hspace{12pt} n\cdot2\hspace{24pt}}{\frac{1}{2}\cdot2\cdot\frac{1}{2}\cdot2\cdot\frac{3}{2}\cdot2\cdot\frac{3}{2}\cdot2\cdot\frac{5}{2}\cdot2\cdot\frac{5}{2}\cdot2\cdot\dots\ (n-\frac{1}{2})\cdot2\cdot(n-\frac{1}{2})\cdot2\cdot(2n+1)}\cdots $$

$$=\frac{1\cdot1\cdot2\cdot2\cdot3\cdot3\cdot \hspace{12pt} \dots\hspace{12pt}n\hspace{12pt}\cdot\hspace{12pt} n\hspace{24pt}}{\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdot\frac{5}{2}\cdot\dots\ (n-\frac{1}{2})\cdot(n-\frac{1}{2})\cdot(2n+1)}\cdots $$

$$=\big(\frac{1\cdot2\cdot3\cdot\ \dots\ n\hspace{24pt}}{\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdot\dots\ (n-\frac{1}{2})\cdot\sqrt{2n+1}}\cdots\big)^2 $$

$$ = \lim_{n\rightarrow\infty}\;\;\Bigg(\frac{\Gamma{(n+1)}}{\frac{\Gamma{(n+\frac{1}{2})}}{\sqrt{\pi}}}\Bigg)^2\frac{1}{(2n+1)}$$

$$=\lim_{n\rightarrow\infty}\;\; \frac{\pi}{2n+1}\Big(\frac{\Gamma{(n+1)}}{\Gamma{(n+\frac{1}{2})}}\Big)^2$$

$$= \frac{\pi}{2}\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2.$$

Question 1) Is this a correct fully equivalent reformulation of rhs of WP?

When using WP we would get:

$$\frac{\pi}{2} = \frac{\pi}{2}\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2,$$

$$\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2 = 1,$$ which implies

$$\frac{\Gamma(n)}{\Gamma(n-\frac{1}{2})} \overset{n\rightarrow \infty}{\longrightarrow} \sqrt{n}. \tag{1}$$

Alternatively if (1) is known otherwise (e.g. from the Sterling approximation (Question 2) Is it possible to derive (1) from the Sterling approximation or from any other elementary considerations?)) one could derive the WP easily from that.

Question 3) Is this conclusion correct?

Raphael J.F. Berger
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2 Answers2

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Let's say $\rho(n) = \Gamma(n)/\Gamma(n-1/2).$ Then, yes, $\rho(n) \approx \sqrt n$ for large $n$. More precisely, $\rho(n)/\sqrt n \to 1$ as $n \to \infty$.

You should expect this. After all, $$\rho(n)\cdot \rho(n-1/2) = \frac{\Gamma(n)}{\Gamma(n-1/2)} \cdot \frac{\Gamma(n-1/2)}{\Gamma(n-1)} = \frac{\Gamma(n)}{\Gamma(n-1)} \approx n$$

for large $n$, so if $\rho(n) \approx \rho(n-1/2)$ for large $n$, then we must have $\rho(n) \approx \sqrt n$ for large n.

Indeed, this observation is one way to pick out the Gamma function as the "best" generalization of factorial to non-integer arguments. For any fixed $t$ between $0$ and $1$ this sort of heuristic argument suggests that you want $$\Gamma(n+t) \approx n^t \cdot \Gamma(n) $$ for large integer $n$. Combined with the functional equation $\Gamma(x+1) = x \cdot \Gamma(x)$, that means for large $n$ you want $$(n-1+t)\cdot\dots\cdot(1+t) \cdot \Gamma(1+t) \approx n^t \cdot (n-1)\cdot \dots \cdot 2\cdot 1.$$

And it turns out that taking the definition $$\Gamma(1+t) = \lim_{n\to \infty} n^t \frac{(n-1)\cdot \dots \cdot 2\cdot 1}{(n-1+t)\cdot\dots\cdot(2+t)\cdot(1+t)}$$ gives you the same Gamma function that we know and love.

This is closely related to the observation that the Gamma function is the unique log-convex extension of the factorial function which satisfies the right functional equation.

Ash Malyshev
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  • OK! So the WP can be directly derived from the asymptotics of $\frac{\Gamma(n)}{\Gamma(n-\frac{1}{2})}$. – Raphael J.F. Berger Aug 11 '17 at 19:56
  • Yes, that answers the second part of the question. The first part ("Is this correct?"/"Can that be right?") is still open. – Raphael J.F. Berger Aug 12 '17 at 06:08
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    Once you know that $\frac{\Gamma(n)}{\Gamma(n-1/2)} = \sqrt{n}(1+o(1))$ as $n \to \infty$, you also know it is correct that $\frac{\pi}{2n+1}\left(\frac{\Gamma(n+1)}{\Gamma(n+1/2)}\right)^2 = \frac{\pi}{2}(1+o(1))$ as $n \to \infty$. So yes, the claim you are asking about is correct. If you're asking about the derivation you presented of that claim, then it is sloppy but probably not wrong in any essential way. That is, it could probably be cleaned up into a correct argument. – Ash Malyshev Aug 12 '17 at 12:48
  • How would that cleaned up correct argument look like? – Raphael J.F. Berger Aug 12 '17 at 12:49
  • You would explicitly write down the $n$'th partial product of the Wallis product (not just the third or the seventh), and manipulate that expression into something that looks more or less like $\frac{\pi}{2n+1} \left( \frac{\Gamma(n+1)}{\Gamma(n+1/2)} \right)^2$, justifying every step of the manipulation. – Ash Malyshev Aug 12 '17 at 12:56
  • Could you include your comment that it is correct in your answer, then I can sign it as answered. – Raphael J.F. Berger Aug 12 '17 at 15:01
  • I made the answer a bit more explicit. It's still not an explicit answer to your question (1), but I don't want to spend time chasing edits to the question with edits to the answer. – Ash Malyshev Aug 12 '17 at 16:32
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By Stirling,

$$\frac{\Gamma(n)}{\Gamma(n-\frac12)}\approx\frac{\sqrt{2\pi n}\left(\dfrac ne\right)^n}{\sqrt{2\pi (n-\frac12)}\left(\dfrac {n-\frac12}e\right)^{n-\frac12}}.$$

After simplifications,

$$\frac1{\left(1-\dfrac1{2n}\right)^n}\left(\frac {n-\frac12}e\right)^{1/2}\to e^{1/2}\sqrt {\frac ne}.$$