# Partial exponential sum

1. Dec 15, 2008

### Nurdan

It is known that
$\sum\limits_{k = 0}^\infty {\frac{{N^k }} {{k!}}} = e^N$

My question is

$\sum\limits_{k = 0}^M {\frac{{N^k }} {{k!}}} = ?$
where $M\leq N$ an integer.

This is not an homework

2. Dec 15, 2008

### djeitnstine

3. Dec 15, 2008

### Tedjn

I do not understand the question. Whether $M \leq N$ or not, as long as M is not infinity, all you have is a Taylor polynomial approximation of eN. The larger M is, the better your approximation.

Your LaTeX is fine, but the forum requires you to put code in between [ tex ][ /tex ] tags (no spaces) or [ itex ][ /itex ] tags (no spaces) for inline TeX.

4. Dec 15, 2008

You should use [tex ][/tex] tags.
What kind of answer are you looking for? There isn't really a simpler expression, and why the restriction M ≤ N?

5. Dec 15, 2008

### Nurdan

My empiric results show that if M=N the answer is (e^N)/2. I am looking for any asymptotic approach that gives the solution as M=N.

6. Dec 15, 2008

Yes, it does seem to approach that (although of course it's not exact).

Interestingly, according to Mathematica,
$$\sum_{k = 0}^{n} \frac{x^k}{k!} = e^x \frac{\Gamma(n + 1, x)}{\Gamma(n + 1)}$$
where $$\Gamma(a, x)$$ is the incomplete gamma function
$$\Gamma(a, x) = \int_x^\infty t^{a - 1} e^t \,dt$$
and $$\Gamma(a) = \Gamma(a, 0)$$.

7. Dec 15, 2008

### Nurdan

It is ok. I know the proof of this but still i need some approximations