# Binomial-like summation

This is a bit embarrassing, but I'm struggling to evaluate this sum.

\begin{equation}\sum_{m=0}^{2k} \frac{x^{m}}{(2k-m)!}\end{equation}

The coefficients are nearly the binomial coefficients, but with an $m!$ missing. (The $(2k)!$ could just be added by hand).

Any pointers?

• Have you tried summing backwards, i.e., changing the iterator from m to 2k - n ? – Lucian Oct 24 '16 at 5:02

## 2 Answers

Consider the series in the following way: \begin{align} \sum_{m=0}^{2k} \frac{x^{m}}{(2k-m)!} &= \frac{x^{0}}{(2k)!} + \cdots + \frac{x^{2k-1}}{1!} + \frac{x^{2k}}{0!} \\ &= \sum_{m=0}^{2k} \frac{x^{2k-m}}{m!} \\ &= x^{2k} \, e_{2k}\left(\frac{1}{x}\right), \end{align} where the $e_{n}(x)$ is the finite exponential functions given by $$e_{n}(x) = \sum_{k=0}^{n} \frac{x^{k}}{k!}.$$

A too long comment, I guess.

Starting from Leucippus's answer, the result could simplify using the incomplete gamma function, leading to

$$\sum_{m=0}^{2k} \frac{x^{m}}{(2k-m)!}=\sum_{m=0}^{2k} \frac{x^{2k-m}}{m!}= e^{\frac{1}{x}} \frac{x^{2 k} }{(2 k)!}\,\Gamma \left(2 k+1,\frac{1}{x}\right)$$