TITLE: Does anyone know what the series $\frac{n^x}{n!}$ converges to?
QUESTION [6 upvotes]: Most people in math are familiar with the conventional taylor series,
$$\sum_{n=0}^{\infty} \frac{x^{n}}{n!}$$ which converges to $$e^x$$
I came across an uncommon series from working with very unusual exponents, and even though it looks simple, I'm not sure how to approach it or much less its result. This one has k in the base in the numerator instead. I am fairly certain it converges, but to what I am not sure, it's not really the same as dealing with x^n. For this specific problem, I don't need to know the method, I am wondering only what it converges to. 
$$\sum_{n=0}^{\infty} \frac{n^{x}}{n!} = ?$$
Does anyone know what this converges to?

REPLY [5 votes]: As also noted by @bof, this is known as Dobiński's formula, i.e. for $x\in \mathbb{N}$, $$B_n = \frac{1}{e}\sum_{k=0}^{\infty} \frac{k^x}{k!}$$ where $B_n$ is the $n$th Bell number. These numbers are catalogued as A000110 in the OEIS. However, for $x\notin \mathbb{N}$, the series $\sum_{k=0}^{\infty} \frac{k^x}{k!}$ is unlikely to have a closed form, although we can clearly see that the series will converge for all $x\in \mathbb{R}$ using the ratio test: $$\lim_{k\to \infty} \left\lvert \frac{k!(k+1)^x}{(k+1)!k^x}\right\rvert = \lim_{k\to \infty} \frac{(k+1)^{x-1}}{k^x} = \lim_{k\to \infty} \frac{1}{k+1}\left(\frac{k+1}{k}\right)^x = 0$$