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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 $\sum\limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$

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Joris Bierkens
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  • That particular estimate is also valid if $a > 1$, just multiply both numerator and denominator by $-1$. – Joris Bierkens Jan 05 '17 at 22:03
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    Using Euler Beta function one gets $\displaystyle (n+1)!\int\limits_0^1 (1-x)^{n-k}\frac{(ax)^{k+1}-(1-x)^{k+1}}{(a+1)x-1}dx$ for the sum. Maybe you can use this for a better approximation. – user90369 Jan 06 '17 at 07:21

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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

MR_BD
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