Asymptotic formula for $\sum_{k=N}^\infty \frac{x^k}{k!}$ as $N \to \infty$

Question

This seems like a weird question, because the series has good convergence and there's no need to use other methods to estimate it for $N \to \infty$.

However, after seeing this question, I tried to come up with some approximation which could allow us to treat integrals containing this sum in the denominator.

Regardless of the particular application, I just wanted to ask if there's a way to approximate this sum using a finite combination of elementary or special functions?

Not counting the obvious $e^x-\sum_{k=0}^{N-1} \frac{x^k}{k!}$ of course.

Euler-Maclaurin formula doesn't seem very promising, because the resulting integral is even more complicated $$\int_N^\infty \frac{x^y}{\Gamma(y+1)}dy$$ And the derivatives containing various combinations of polygamma functions are also a pain to write.

If there's no better asymptotic expression than the sum itself, so be it.

Answer 1

Note that $$ \frac{n!}{x^n}\sum_{k=n}^\infty\frac{x^k}{k!} =1+\sum_{k=n+1}^\infty\frac{x^{k-n}n!}{k!}\\ $$ and for $n\ge|x|$, $$ \begin{align} \left|\sum_{k=n+1}^\infty\frac{x^{k-n}n!}{k!}\right| &\le\sum_{k=1}^\infty\frac{|x|^k}{(n+1)^k}\\ &=\frac{|x|}{n+1-|x|} \end{align} $$ Thus, $$ \sum_{k=n}^\infty\frac{x^k}{k!} =\frac{x^n}{n!}\left(1+O\!\left(\frac1n\right)\right) $$

Answer 2

In another way, note that your sum is related with the Regularized Incomplete Gamma Function. $$ \eqalign{ & Q(s,z) = {{\Gamma (s,z)} \over {\Gamma (s)}} = e^{-z} \sum_{k = 0}^{s} {{{z^{k}} \over {k!}}} = \left( {1 - e^{-z} \sum_{k=s}^{\infty } {{{z^k} \over {k!}}} } \right)\quad \Longrightarrow \cr & \Longrightarrow \quad \sum _{k=s}^{\infty } {{{z^{k}} \over {k!}}} = e^{z} P(s,z) = e^{z} {{\gamma (s,z)} \over {\Gamma (s)}} \cr} $$ note that here the summation is taken in the Antidelta meaning.

Therefore, using the asymptotic expansion of $Q(s,z)$ for $s \, \to \, \infty$ we get $$ \sum_{k=s}^{\infty } {{{z^k} \over {k!}}} = \sum_{k = s}^{\infty } {{{z^k} \over {k!}}} = e^{z} P(s,z) = {{s^{- s - 1/2} e^{s} z^{s} } \over {\sqrt {2\pi } }}\left( {1 + {{12z - 1} \over {12s}} + O\!\left( {{1 \over {s^{\,2} }}} \right)} \right),\quad \left| s \right| \to \infty. $$

Answer 3

Maple tells me it's $$\dfrac{x^n}{n!} \left(1 + \frac{x}{n} + \frac{x(x-1)}{n^2} + \frac{x(x^2-3x+1)}{n^3} + \frac{x(x^3 - 6x^2 + 7x - 1)}{n^4} + O(1/n^5)\right)$$

You can get this by a similar method to robjohn's.