Infinite series involving e

Question

Consider: $$f(x) = \frac{1}{e}\sum_{n=0}^{\infty}\frac{n^x}{n!}$$ When $x$ is a whole number f(x) is also a whole number. Is there a way to resolve this summation?

Answer 1

One method of examination is to obtain a generating function. For this is can be determined that: \begin{align} \sum_{n=0}^{\infty} f(n) \, \frac{t^n}{n!} &= \sum_{n=0}^{\infty} \frac{t^{n}}{n!} \, \frac{1}{e} \, \sum_{k=0}^{\infty} \frac{k^{n}}{k!} \\ &= \frac{1}{e} \, \sum_{k=0}^{\infty} \frac{1}{k!} \, \sum_{n=0}^{\infty} \frac{(k t)^{n}}{n!} \\ &= \frac{1}{e} \, \sum_{k=0}^{\infty} \frac{e^{k t}}{k!} \\ &= e^{e^{t} - 1}. \end{align} This is the generating function for the Bell numbers, $B_{n}$. This implies that $f(n) = B_{n}$.