How Accurate Are Partial Sums in Estimating e^N?

[Nurdan]

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


I am looking for any asymptotic approximation which gives

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


This is not an homework

 

[christianjb]

I'm only being a little facetious if I point out that the sum is asymptotically equal to e^N.

Want more accuracy? It's better approximated by e^N-[x^(N+1)]/(N+1)!