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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?

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    This is not, strictly speaking, a Taylor series. I don't know that it has a closed-form expression. – Michael L. Jul 30 '17 at 03:28
  • It is unusual isn't it? I typed it into a computation engine and it does converge apparently, but it again didn't define what it converged to. This would suggest it does have a closed form, but that it is a function that is particularly difficult to calculate or one that is not currently defined in terms of known functions. –  Jul 30 '17 at 03:30
  • Since the result has something to do with a series of exponentials, would it be possible to define it as some kind of Euler product? Or perhaps it is part of some kind of generalized hypergeometric series? –  Jul 30 '17 at 03:40
  • As far as I know that series has not a closed form. – Ixion Jul 30 '17 at 03:40
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    No, convergence doesn't necessarily suggest a closed form. I would venture that the vast majority of functions on $\mathbb{R}$ are not expressible in terms of elementary functions, or even in terms of already-defined functions. However, for $x\in \mathbb{N}$, the series converges to $e\cdot $A000110(x). – Michael L. Jul 30 '17 at 03:40
  • @MichaelLee Sorry, but could you type the actual latex form what you found the result to be so that we can see it as a nice expression here like in the original post? The denominator is a factorial, and factorials grow faster than exponents, so I don't see immediately why it wouldn't converge. –  Jul 30 '17 at 03:41
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    For $x\in \mathbb{N}$, $\sum_{n=1}^{\infty} \frac{n^x}{n!} = A_xe$, where $A_x$ is the $x$th term of A000110. – Michael L. Jul 30 '17 at 03:43
  • Indexing starts at $0$. – Michael L. Jul 30 '17 at 03:44
  • It converges, but convergence doesn't suggest a closed form. As is apparent from the existing writing on this sequence, there is nothing to suggest an easy closed form thus far. – Michael L. Jul 30 '17 at 03:45
  • @MichaelLee You beat me to it! – bof Jul 30 '17 at 03:46
  • @MichaelLee That's fair enough, I mean for this specific problem, since it seems apparent that factorials grow faster than these exponents, that you can assume it converges to something. But anyway, so you're saying it's a Bell number, and you're sure you mean the $$nth$$ term and not the $$xth$$? Because the "nth" term would suggest it's still somehow locked in the summation after the summation has been calculated, or that it is a partial summation. –  Jul 30 '17 at 03:46
  • Of course, yes. Hasty typo on my part. – Michael L. Jul 30 '17 at 03:48
  • @MichaelLee Alright just making sure, thank you for answering. Oh hey, there we go, "bell numbers are given in terms of generalized hypergeometric functions" http://mathworld.wolfram.com/BellNumber.html –  Jul 30 '17 at 03:48
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    For natural $x$ this is Dobiński's formula. – bof Jul 30 '17 at 03:51
  • @MichaelLee Could you put your answer is as a question answer so I can put that check mark that says this question's answered? –  Jul 30 '17 at 04:44
  • This is an ok question. Please give some tips to your roommate. – Jyrki Lahtonen Jul 30 '17 at 06:27

1 Answers1

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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: $$\lim_{k\to \infty} \left\lvert \frac{k!(k+1)^x}{(k+1)!k^x}\right\rvert = \lim_{k\to \infty} \frac{(k+1)^{x-1}}{k^x} = \lim_{k\to \infty} \frac{1}{k+1}\left(\frac{k+1}{k}\right)^x = 0$$

bof
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Michael L.
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