# Convergence of a sequence involving a truncated exponential

Let $n\in\mathbb{N}_{>0}$, $\gamma\in\mathbb{R}_{>0}$. Let $\{a_n\}_{n}$ and $\{b_n\}_{n}$ be two sequences defined as follows $$a_n := \sum_{k=0}^{n-1}{2k \choose k}\frac{1}{\gamma^{2k+1}} \left(1-e^{-\frac{\gamma}{n}}\sum_{h=0}^{2k} \frac{(\gamma/n)^h}{h!} \right),$$ $$b_n := \sum_{k=0}^{n-1}{2k \choose k}\frac{1}{\gamma^{2k+1}} e^{-\frac{\gamma}{n}}\sum_{h=0}^{2k} \frac{(\gamma/n)^h}{h!}.$$ Moreover, let $$c_n:=\left(\frac{a_n}{b_n}\right)^n.$$

My question. What can be said about the convergence of the sequence $\{c_n\}_{n}$?

N.B. This is not an homework. It is part of a problem that came up in my research some time ago. Experimentally, I observed the $\{c_n\}_{n}$ converges to a finite non-zero value for any (randomly selected) $\gamma>0$. Thus, I wonder whether there is a chance to prove this.