What is the sum of the following series? $k$ is a natural number.
$\sum_{n\geq 1}\frac{n^k}{n!}$
It appears to always provide an integer multiple of $e$.
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1Are you asking what the sum of the infinite series is? What do you mean the series is an answer? What work have you done on the problem, and where are you stuck? What is the context of the problem? – Rory Daulton Apr 30 '15 at 11:53
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Yes, it always provide a integer multiple of $e$, and more, the secuence of those integers is called Bell numbers. – luisfelipe18 Apr 30 '15 at 12:10
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This same question appeared at this MSE link. – Marko Riedel Apr 30 '15 at 19:47
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Let $$A(k) = \sum_{n\geq 1}\frac{n^k}{n!}.\tag{1}$$ Since: $$\sum_{n\geq 1}\frac{e^{nx}}{n!} = e^{e^x}-1 \tag{2}$$ we have: $$ A(k) = \frac{d^k}{dx^k}\left.\left(e^{e^x}-1\right)\right|_{x=0} = B_k\cdot e\tag{3}$$ where $B_k$ is a Bell number. $(3)$ is also known as Dobinski's formula.
Jack D'Aurizio
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5This is a surprisingly interesting answer for such a terrible question. – theage Apr 30 '15 at 11:59