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I am trying to find a general formula for the following summation;

$$S_k=\sum_{n=0}^∞ \dfrac{n^k}{n!}$$ So naturally, first I tried to find the expression for $k=1,2,3,4$ to see if I can get a pattern and it appears as though $S_k=(2k-1)e$ where $e$ is Napier's constant;

$$S_1=\sum_{n=0}^∞ \dfrac{n}{n!}=e$$ $$S_2=\sum_{n=0}^∞\dfrac{n^2}{n!}=\sum_{n=0}^∞\dfrac{n-1}{(n-1)!}+\sum_{n=0}^∞\dfrac{2}{n!}=3e$$ Similarly, for $k=3$, we have; $$S_3=\sum_{n=0}^∞\dfrac{n^2-1}{(n-1)!}+e=\sum_{n=0}^∞\dfrac{n+1}{(n-2)!}+e=\sum_{n=0}^∞\dfrac{1}{(n-3)!}+4e=5e$$

In all the above computations, I'm using the fact that negative factorials are infinity so their reciprocal is $0$ and has no effect on the sum. And of course, that $e=\sum_{n=0}^∞ \frac1{n!}$

However, I'm unable to prove this assertion. I tried to set up a recursion in $S_k$ and prove this formula via induction but I do not know how to solve this kind of recursion; it somewhat resembles a formula for an arithmetic progression whose differences are themselves in an arithmetic progression I'd seen earlier but I'm unable to correlate it well.

$$S_k=\sum_{n=0}^∞ \dfrac{n^k}{n!}=S_k=\sum_{n=0}^∞ \dfrac{n^{k-1}-1}{(n-1)!}+e$$ $$\implies S_k=\sum_{n=0}^∞ \dfrac{n^{k-2}+n^{k-3}+...1}{(n-2)!}+e=\sum_{i=1}^{k-2} S_i +2e$$

Can anyone help me solve this problem?

Cognoscenti
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  • See https://math.stackexchange.com/questions/2733380/value-of-sum-n-0-infty-fracnkn-for-k-in-mathbbn?noredirect=1. The real answer is in the comments. – X-Rui Feb 02 '24 at 14:36
  • I believe there is a recurrence formula expressing S_k in terms of S_j for j < k, but I have no references to suggest. – Dan Asimov Feb 02 '24 at 14:39
  • You can build a generating function for $k=0, 1, 2...$ $$S_k=\sum_{n=0}^\infty \frac{n^k}{n!}=\frac{d^k}{d\alpha^k},\bigg|{\alpha=0}\sum{n=0}^\infty \frac{e^{\alpha n}}{n!}=\frac{d^k}{d\alpha^k},\bigg|_{\alpha=0}e^{e^\alpha}$$ – Svyatoslav Feb 02 '24 at 14:44
  • Yeah, okay I got it, I'm sorry this is a complete duplicate, I didn't find the other answers before, thank you for directing me to them – Cognoscenti Feb 02 '24 at 15:49

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The problem could be solved if we could find numbers $ a_{n,k} $ such that we would have : $ x^{n}=\sum\limits_{k=1}^{n}{a_{n,k}x\left(x-1\right)\cdots\left(x-k+1\right)}$. For example we have in particular : \begin{aligned} x^{2}&=x\left(x-1\right)+x\\ x^{3}&=x\left(x-1\right)\left(x-2\right)+3x\left(x-1\right)+x\\ x^{4}&=x\left(x-1\right)\left(x-2\right)\left(x-3\right)+6x\left(x-1\right)\left(x-2\right)+7x\left(x-1\right)+x \end{aligned}

Such numbers are called Stirling numbers of the second kind. written $ \left\lbrace n\atop k\right\rbrace$, $ 1\leq k\leq n$.

Forall $ n,k\geq 1 $, we have the following :

$$ n^{k}=\sum_{p=1}^{k}{\left\lbrace k\atop p\right\rbrace n\left(n-1\right)\cdots\left(n-p+1\right)} $$

Thus :\begin{aligned}\sum_{n=0}^{+\infty}{\frac{n^{k}}{n!}}&=\sum_{p=1}^{k}{\left\lbrace k\atop p\right\rbrace\sum_{n=0}^{+\infty}{\frac{n\left(n-1\right)\cdots\left(n-p+1\right)}{n!}}}\\ &=\sum_{p=1}^{k}{\left\lbrace k\atop p\right\rbrace\sum_{n=p}^{+\infty}{\frac{1}{\left(n-p\right)!}}}\\ &=\mathrm{e}\sum_{p=1}^{k}{\left\lbrace k\atop p\right\rbrace}\end{aligned}

CHAMSI
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