$$\displaystyle \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{2} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{2} + 1\right)}}$$

This is from here.

If we add one more variable -

$$\displaystyle L(p,q,m,t) = \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{t} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{t} + 1\right)}}$$

Then, the results I got from W|A are really interesting. They seem to follow the pattern which says

$\displaystyle L(p,q,m,t) = \dfrac{1}{t^{|p-q|/m}}$

If it is indeed true, how to prove it? Factorial approximations did not lead me anywhere.

  • $\begingroup$ Have you tried taking logarithms and then applying Stirling's formula? $\endgroup$ – Dominik Dec 7 '16 at 10:56
  • $\begingroup$ What Dominik suggests works like a charm. $\endgroup$ – Claude Leibovici Dec 7 '16 at 11:58

In the same spirit as Dominik's answer, consider $$A=\displaystyle L(p,q,m,t) = \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{t} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{t} + 1\right)}}$$ and take logarithms.

Now use Stirling approximation $$\log\left( \Gamma\left(x\right)\right)=x (\log (x)-1)+\frac{1}{2} \left(\log (2 \pi )-\log(x)\right)+O\left(\frac{1}{x}\right)$$ Apply to each term and simplify as much as you can.

Now, apply to the result Taylor again for large values of $n$ to get (to second order)

$$\log(A)=\frac{(q-p) \log \left(\frac{1}{t}\right)}{m}+\frac{ (q-p) (m+p+q)(t-1)}{2 m^2 n}+O\left(\frac{1}{n^2}\right)\tag 1$$ which makes the limit to be $$\log(A)=\frac{(q-p) \log \left(\frac{1}{t}\right)}{m}=\frac{(p-q) \log \left(t\right)}{m}\implies A= L(p,q,m,t)={t^{\frac{p-q}{m}}}$$ The second term in $(1)$ shows how the limit is approached.


From Abramowitz and Stegun we know $$\ln \Gamma(x) = x \ln(x) - \frac{1}{2} \ln(x) - x + \frac{\ln(2\pi)}{2} + O(\tfrac{1}{x}) \qquad (x \to \infty)$$

Applying this formula here yields

$$\begin{align*} \ln \Gamma\left(\frac{p}{m} + \frac{n}{t}\right) &= \left(\frac{p}{m} + \frac{n}{t}\right) \ln\left(\frac{p}{m} + \frac{n}{t}\right) - \left(\frac{p}{m} + \frac{n}{t}\right) + \frac{\ln(2\pi)}{2} + O(\tfrac{1}{n}) \\ &= \left(\frac{p}{m} + \frac{n}{t}\right) \ln(n) - \frac{n}{t} - \frac{p}{m} + \left(\frac{p}{m} + \frac{n}{t}\right) \ln \left(\frac{1}{t} + \frac{p}{mn}\right) + \frac{\ln(2\pi)}{2} + o(1). \end{align*}$$

After cancelling some terms, we get $$\ln \frac{\Gamma\left(\frac{p}{m} + \frac{n}{t}\right)}{\Gamma\left(\frac{q}{m} + \frac{n}{t}\right)} = \frac{p-q}{m} \ln(n) - \frac{p - q}{n} + \frac{p - q}{m} \ln\left(\frac{1}{t}\right) + \frac{n}{t} \ln \left(\frac{\frac{1}{t} + \frac{p}{mn}}{\frac{1}{t} + \frac{q}{mn}}\right) + o(1).$$

Now note that $$\frac{n}{t} \ln \left(\frac{\frac{1}{t} + \frac{p}{mn}}{\frac{1}{t} + \frac{q}{mn}}\right) = \frac{1}{t} \ln\left(\frac{\left(1 + \frac{pt}{mn}\right)^n}{\left(1 + \frac{qt}{mn}\right)^n}\right) = \frac{1}{t}\ln\left(\frac{\exp(pt/m)}{\exp(qt/m)}\right) + o(1) = \frac{p - q}{m} + o(1).$$

After cancelling some more terms we get $$\ln \frac{\Gamma\left(\frac{p}{m} + n\right) \Gamma\left(\frac{q}{m} + \frac{n}{t}\right)}{\Gamma\left(\frac{q}{m} + n\right)\Gamma\left(\frac{p}{m} + \frac{n}{t}\right)} = \frac{q - p}{m} \ln(\tfrac{1}{t}) + o(1) = \frac{p - q}{m} \ln(t) + o(1).$$

Using the formula $\Gamma(x + 1) = x \Gamma(x)$ we see that the limit in the last equation is equal to $\ln L(p, q, m, t)$, which finally shows

$$L(p, q, m, t) = t^{(p-q)/m}.$$

  • $\begingroup$ Your final equation should have a mod. It looks ambiguous like this. $\endgroup$ – Kartik Sharma Dec 7 '16 at 17:24
  • $\begingroup$ @KartikSharma What do you mean? I don't see how the formula could be ambiguous. $\endgroup$ – Dominik Dec 7 '16 at 17:25
  • $\begingroup$ The original limit is symmetric for $p,q$ but your formula is not symmetric. $\endgroup$ – Kartik Sharma Dec 7 '16 at 17:34
  • $\begingroup$ That's because the original formula is incorrect. It is also pretty obvious that it is incorrect in this way, since $L(p, q, m, t) = L(q, p, m, t)^{-1}$. $\endgroup$ – Dominik Dec 7 '16 at 17:41
  • 1
    $\begingroup$ I have mixed up $p$ and $q$ near the end. Now the result should be correct. $\endgroup$ – Dominik Dec 8 '16 at 17:55

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