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I am trying to find a closed form expression for $$S_k = \sum_{n=0}^{\infty} \frac{n^k}{n!}$$

For $k=1$, the answer is clearly $e$. For $k=2$, $$\sum_{n=0}^{\infty} \frac{n^2}{n!} = \sum_{n=1}^{\infty} \frac{n^2}{n!} = \sum_{n=1}^{\infty} \frac{n}{(n-1)!} = \sum_{n=0}^{\infty} \frac{n+1}{n!} = S_1 + e = 2e$$

Following the same reduction steps for $n=3$, I get $5e$ ($S_2 + 2S_1 + e$)

Using my computer to find the next few values, the terms seem to be $e, 2e, 5e, 15e, 52e,203e, 877e, ...$

I don't see any discernible pattern in this, otherwise I could solve this by induction using the reduction steps outlined for $S_2$. Alternatively, if we denote $S_0$ to be $e$:

$$\sum_{n=0}^{\infty} \frac{n^k}{n!} = \sum_{n=1}^{\infty} \frac{n^k}{n!} = \sum_{n=1}^{\infty} \frac{n^{k-1}}{(n-1)!} = \sum_{n=0}^{\infty} \frac{(n+1)^{k-1}}{n!} = \sum_{n=0}^{\infty} \frac{\sum_{r=0}^{k-1} \binom{k-1}{r} n^r}{n!} = \sum_{r=0}^{k-1} \binom{k-1}{r} S_r$$

So $S_k$ follows the recurrence relation: $$S_k = \sum_{r=0}^{k-1} \binom{k-1}{r} S_r$$

I don't know how to solve this recurrence either, because it has a variable number of terms for each next term unlike a linear recurrence.

How can I find the closed form for $S_k$? Does it even exist? I don't really know much about these kinds of sequences (I'm in high school) so if you could also provide some sources where I can look for more information on these kinds of series, that would me much appreciated.

  • Also: https://math.stackexchange.com/q/1266331/42969, https://math.stackexchange.com/q/3672792/42969 – Martin R Feb 24 '23 at 19:18

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A great resource for questions like this is the Online Encyclopedia of Integer Sequences. If you search the first few terms of your sequence there, you'll find that the numbers $1,2,5,15,52,\dots$ are the Bell numbers. There are some combinatorial descriptions, representations in terms of other sequences, et cetera. Depending on what you'd consider a "closed form," I'm not sure you'll find anything there, which probably means there isn't one (you can't represent it as a finite expression in terms of binomial coefficients, for example), but there's given a relation between these number and Stirling numbers of the second kind $$B_n=\sum_{k=0}^n \left\{n\atop k\right\},$$ which may be of interest. (I believe these $B_n$ are $e^{-1}S_n$, in your notation.) The encyclopedia also lists the definition you gave, the recurrence you derived, the potentially useful expression $$\sum_{n=0}^\infty \frac{B_nx^n}{n!}=e^{e^x-1}$$ (this is the exponential generating function), and much more.