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 $k$.



Infinite limit of sums of gamma functions is constant?
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 $\alpha\rightarrow\infty$.


