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Why is it true that $$\lim_{n\to\infty} \frac{\Gamma\left(n - \frac{1}{2}\right)}{\Gamma\left(n\right)} = e^{-\frac{1}{2}}$$

I only know the integral definition of gamma function. My notes writes $$\lim_{n\to\infty} \frac{\Gamma\left(n - \frac{1}{2}\right)}{\Gamma\left(n\right)} = \lim_n\prod_{k=1}^{n-1}\left(1 - \frac{1/2}{k}\right) = e^{-1/2}$$

I don't know why the first equality holds, nor why the second equality holds...

3x89g2
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    The second equality definitely does not hold. $\sum_k \frac{1}{2k} = +\infty \implies \prod_{k=1}^\infty (1-\frac{1}{2k}) = 0$ – mathworker21 Jul 28 '18 at 03:34
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    Is it even true? W.A. seems to say otherwise (although I know WA isn't always correct): http://www.wolframalpha.com/input/?i=limit+of+gamma(n-0.5)%2Fgamma(n) – Brenton Jul 28 '18 at 03:35
  • @Brenton I am not sure, to be honest. This note could be wrong. Anyway, I still need to evaluate the limit but not sure how to. My guess was that the limit would be $1$ (both $\Gamma(n-1/2)$ and $\Gamma(n)$ grows at similar rate?) but don't know if it's true or not. – 3x89g2 Jul 28 '18 at 03:38
  • As for why you should expect it to be zero: what is the limit of $(n - 1)! / n!$? (Then you can compare $ \frac{ \Gamma( n - 1/2 - 1/2) }{ \Gamma(n - 1/2) } \frac{ \Gamma( n - 1/2) }{ \Gamma(n)}= \Gamma(n - 1)/\Gamma(n) = (n-2)!/(n-1)!$, which you can turn into a proof, provided you can argue that the limit exists...) – Elle Najt Jul 28 '18 at 03:53
  • To get some intuition, notice that $$\frac{\Gamma(n-k)}{\Gamma(n)}=\frac{1}{(n-1)(n-2)\cdots(n-k)}\sim \frac{1}{n^k}\quad\text{as }n\to\infty$$ for each positive integer $k$. So we can expect that a similar behavior persists for non-integral $k$. Indeed, this extrapolation is justified by the Stirling's approximation. – Sangchul Lee Jul 28 '18 at 04:05

3 Answers3

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This is not true. If it were true, we'd have

\begin{eqnarray*} \lim_{n\to\infty}\frac{\Gamma\left(n-1\right)}{\Gamma(n)} &=& \lim_{n\to\infty}\left(\frac{\Gamma\left(n-\frac12\right)}{\Gamma(n)}\cdot\frac{\Gamma(n-1)}{\Gamma\left(n-\frac12\right)}\right) \\ &=& \lim_{n\to\infty}\frac{\Gamma\left(n-\frac12\right)}{\Gamma(n)}\cdot\lim_{n\to\infty}\frac{\Gamma(n-1)}{\Gamma\left(n-\frac12\right)} \\ &=& \mathrm e^{-\frac12}\cdot\mathrm e^{-\frac12} \\ &=& \mathrm e^{-1}\;, \end{eqnarray*}

whereas this limit is in fact $0$ (since the quotient is $\frac1{n-1}$).

The correct statement that this might be intended to state might be

$$ \frac{\Gamma\left(n-\frac12\right)}{\Gamma(n)}\sim n^{-\frac12}\;. $$

joriki
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  • May I ask how do you get the third equal sign (How to evaluate those two limits)? If it's approximately $\sqrt{\frac{1}{n}}$ then wouldn't it goes to zero as $n\to\infty$? – 3x89g2 Aug 02 '18 at 03:32
  • @3x89g2: Indeed. It's a proof by contradiction. It assumes the OP's false claim in order to disprove it by deriving.a contradiction from it. It's prefaced with: "This is not true. If it were true, we'd have ...". The third equals sign uses the OP's false claim twice (once directly, once shifted). – joriki Aug 02 '18 at 05:00
  • Ah right... I misread. Thank you. – 3x89g2 Aug 02 '18 at 05:01
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Considering the general case$$f_a=\frac{\Gamma (n-a)}{\Gamma (n)}\implies \log(f_a)=\log (\Gamma (n-a))-\log (\Gamma (n))$$ and use Stirling approximation to get $$\log(f_a)=-a \log \left({n}\right)+\frac{a(a+1)}{2 n}+O\left(\frac{1}{n^2}\right)$$ Continuing with Taylor series $$f_a=e^{\log(f_a)}=n^{-a} \left(1+\frac{a(a+1)}{2 n}+O\left(\frac{1}{n^2}\right) \right)$$

Claude Leibovici
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The first equality is not true reads the comments.

for the second equality cannot be true

note that by telescopic sum we have

$$ \ln\left(\prod_{k=1}^{n-1}\left(1 - \frac{1/2}{k}\right)\right) = \sum_{k=1}^{n-1}\ln\left(2k - 1\right)-\ln\left(2k \right)= \sum_{j=2}^{2n-2}\ln\left(j- 1\right)-\ln\left(j \right)\\ = \ln\left(2- 1\right)-\ln\left(2n-2 \right)\to-\infty $$

Whence

$$\prod_{k=1}^{n-1}\left(1 - \frac{1/2}{k}\right) \to0.$$

Guy Fsone
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    Your summation index change isn't valid. In

    $$ \sum_{k=2}^{2n-2}\ln\left(j- 1\right)-\ln\left(j \right)$$

    you probably meant the summation index to be $j$ rather than $k$; but that sum isn't equal to the previous one; you now have twice as many terms.

     – joriki Jul 28 '18 at 05:50
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    Also, the first equation is not true. $\Gamma\left(n-\frac12\right)$ isn't a product of $n$ factors of the form $k-\frac12$, as appears to be assumed in that equation. – joriki Jul 28 '18 at 05:53