How can I find the value of the sum $\sum_{k=1}^{\infty}\frac{k^n}{k!}$? for example for $n=6$, we have

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    $\begingroup$ @michaelrozenberg, please do not edit many questions in a row, as that makes the from page like a vision of Christmas Past. $\endgroup$ – Mariano Suárez-Álvarez Dec 28 '16 at 5:56

I thought it might be useful to add the derivation of the formula given in Casteels answer.

The defining equation for Stirling Numbers of the Second Kind is $$ \newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}} k^n=\sum_{j=0}^n\stirtwo{n}{j}\binom{k}{j}\,j!\tag{1} $$ Therefore, $$ \begin{align} \sum_{k=0}^\infty\frac{k^n}{k!} &=\sum_{k=0}^\infty\sum_{j=0}^n\stirtwo{n}{j}\binom{k}{j}\frac{j!}{k!}\\ &=\sum_{j=0}^n\stirtwo{n}{j}\sum_{k=j}^\infty\frac1{(k-j)!}\\ &=e\sum_{j=0}^n\stirtwo{n}{j}\tag{2} \end{align} $$ It is worth noting that the Bell Numbers are $$ B_n=\sum_{j=0}^n\stirtwo{n}{j}\tag{3} $$ Thus, $$ \sum_{k=0}^\infty\frac{k^n}{k!}=e\,B_n\tag{4} $$

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    $\begingroup$ Thanks, I'm so glad that see you once again on my questions!!! $\endgroup$ – user91500 Nov 9 '15 at 13:56
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    $\begingroup$ Nice question! It was listed as related to this question, which I answered a while ago. It got some attention recently and so I took a look at some of the related questions. $\endgroup$ – robjohn Nov 9 '15 at 14:01

The answer is $eB_n$, where $B_n$ is the $n$th Bell number. This is known as Dobinski's formula.


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