# Infinite limit of sums of gamma functions is constant?

The following expression arises in the study of hierarchical models. I suspect that the sum of the underlined $$4$$ terms become constant as $$\alpha\rightarrow \infty$$. Mathematica agrees when prompted with 'toy' versions, but I'm having some difficulty seeing how it generalizes.

I would greatly appreciate any help or observations.

\begin{align} \log \text{P}\propto \underbrace{ \log \Gamma(k \alpha) - \log \Gamma(k\alpha + length(\ell)) - k \log \Gamma(\alpha) + \sum^k_{i=1} \log \Gamma(c_{i} + \alpha)} \ + \ f(\text{other parameters}) \end{align} Where: $$\ell$$ is a list of items; Each item in $$\ell$$ must be assigned to exactly one bin, so $$c_{i}$$ is the count of items in $$\ell$$ assigned to bin $$i$$, and there are a total of $$k$$ bins.

Current thinking: It is well known that as $$\alpha\rightarrow \infty$$, the argument of the sum in the rightmost term $$\log \Gamma(c_{i}+a)$$ increases only linearly in its arguments rather than superlinearly. This is because $$\displaystyle \lim_{\alpha\rightarrow\infty} \big(\frac{\partial^2}{\partial\alpha^2}\log \Gamma(c_{i} + \alpha)\big) = \lim_{\alpha\rightarrow\infty} \big(\frac{\partial^2}{\partial c_{i}^2}\log \Gamma(c_{i} + \alpha)\big) = \displaystyle \lim_{\alpha\rightarrow\infty} \big(\frac{\partial}{\partial c_{\ell,i}}\Psi(c_{i} + \alpha)\big)=0$$, where $$\Psi$$ is the Digamma function. Therefore any partition of $$\ell$$ (i.e. any choices of the different $$c_{i}$$) will cause this term to sum to the same constant.

The other terms are all constant for fixed $$\alpha$$, $$\ell$$, and $$k$$.

• in the third term, is the argument of the Gamma function $k\alpha+\text{length}(\ell)$ or is it $k\alpha+k\,\text{length}(\ell)$ ? Feb 23, 2023 at 8:49
• Thank you -- question edited to clarify (that term's argument is (ππΌ+πππππ‘β(β)). Feb 23, 2023 at 9:31

For each $$k>0$$, $$c\in\mathbb{C}$$ it holds that $$\lim_{\alpha\rightarrow\infty}\frac{\Gamma(\alpha+c)}{\Gamma(\alpha)\alpha^c}=1\Rightarrow\lim_{\alpha\rightarrow\infty} \left(\log\frac{\Gamma(k\alpha+c)}{\Gamma(k\alpha)}-c\log k\alpha\right)=0.$$ Apply this to $$I= - \log \frac{\Gamma(k\alpha + L)}{\Gamma(k\alpha)} + \sum^k_{i=1} \log\frac{\Gamma(\alpha+c_i )}{\Gamma(\alpha)}$$ and you find (using $$\sum_{i=1}^k c_i=L$$) that $$\lim_{\alpha\rightarrow\infty} I=-L\log k\alpha+\sum_{i=1}^k c_i\log\alpha=-L\log k.$$ So indeed, the function $$\log P$$ approaches an $$\alpha$$-independent limit for $$\alpha\rightarrow\infty$$.
• Ahh of course it applies a second time -- thank you! May I ask where you found the very useful property $\lim_{\alpha\rightarrow\infty} \left(\log\frac{\Gamma(k\alpha+c)}{\Gamma(k\alpha)}-c\log k\alpha\right)=0.$? Feb 23, 2023 at 13:25
• this follows from the asymptotics $\Gamma(x+y)\rightarrow\Gamma(x)x^y$ for $x\rightarrow\infty$ at fixed $y\in\mathbb{C}$, see Wikipedia Feb 23, 2023 at 13:37