How Accurate Are Partial Sums in Estimating e^N?

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SUMMARY

The forum discussion focuses on the accuracy of partial sums in estimating the exponential function e^N. It establishes that the infinite series \(\sum\limits_{k = 0}^\infty {\frac{{N^k }}{{k!}}} = e^N\) serves as the foundation for approximating the finite sum \(\sum\limits_{k = 0}^M {\frac{{N^k }}{{k!}}}\) where M is less than or equal to N. The discussion concludes that for improved accuracy, the approximation can be refined to e^N - \(\frac{x^{(N+1)}}{(N+1)!}\).

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Nurdan
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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
 
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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)!
 

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