Let $a$ and $b$ be two positive real numbers and define $$ f(n)=\binom{2n-2}{n-1}\frac{b^{2n-2}}{(2a)^{2n+1}}\left(1-e^{-2a}\sum_{k=0}^{2n-2}\frac{(2a)^k}{k!}\right), \ \ \ n\in\mathbb{N}_{\ge 1}. $$
Question. Numerical simulations seem to suggest that $\lim_{n\to \infty}f(n)\to \infty$ if $b>a$. However, can this fact be proved?
Remark. The function $f(n)$ can be equivalently rewritten in terms of the lower incomplete gamma function $\gamma(\cdot,\cdot)$ as: $$ f(n)=\frac{\gamma(2n-1,2a)}{(n-1)!(n-1)!}\frac{b^{2n-2}}{(2a)^{2n+1}}, \ \ \ n\in\mathbb{N}_{\ge 1}. $$
The limit is $0$ for any $a, b > 0$. This is because $\binom{2n-2}{n-1} \frac{b^{2n-2}}{(2a)^{2n+1}} $ grows exponentially while the factor $1-e^{-2a}\sum_{k=0}^{2n-2} \frac{(2a)^k}{k!}$ decays super-exponentially.
Indeed, write $m = n-1$ and assume that $m$ is large enough to satisfy $2m+1 > 2a$. Then
\begin{align*} 1 - e^{-2a} \sum_{k=0}^{2m} \frac{(2a)^k}{k!} &= e^{-2a} \sum_{k=2m+1}^{\infty} \frac{(2a)^k}{k!} \\ &\leq e^{-2a} \sum_{j=0}^{\infty} \frac{(2a)^{2m+1+j}}{(2m+1)!(2m+1)^j} \\ &= \frac{(2a)^{2m+1}}{(2m)!} \frac{e^{-2a}}{2m+1 - 2a} \end{align*}
Together with the crude bound $\binom{2m}{m} \leq 4^m$, we have
\begin{align*} f(n) &\leq 4^m \frac{b^{2m}}{(2a)^{2m+3}} \frac{(2a)^{2m+1}}{(2m)!} \frac{e^{-2a}}{2m+1 - 2a} \\ &= \frac{1}{4a^2 e^{2a}(2m+1-2a)} \frac{(2b)^{2m}}{(2m)!} \end{align*}
This bound converges to $0$ as $m\to\infty$.