Problem: Simplify $$ \sum_{m=0}^k a^m m! (n-m)!$$ where $a$ is real and positive, $k \in \mathbb N$ and $n \in \mathbb N$ with $k \leq n$.
Upper bounds and lower bounds may also be useful. One example I have found is, assuming $a \neq 1$ and using $m! (n-m)! \leq n!$, $$ \sum_{m=0}^k a^m m!(n-m)! \leq n! \sum_{m=0}^k a^m = \frac{n! \left(1 - a^{k+1} \right)}{1-a},$$ but I am not satisfied with the accuracy of this bound.
I am aware of the somewhat related math.stackexchange.com question: Calculate $\sum\limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$
One Idea is to estimate entropy function based on the value of $k$ (if you know the range of it)
$n \choose m $ $\approx 2^{nH(\frac{m}{n})}$
Now, if you know $k$ you can find a linear approximation to $H(x)$ in $[0,\frac{k}{n}]$ and use it to simply calculate the summation