I recently started studying about the exponential series, and came across this infinite series $ {S}_{k}\mathrm{{=}}\mathop{\sum}\limits_{{n}\mathrm{{=}}{0}}\limits^{\mathrm{\infty}}{\frac{{n}^{k}}{n\mathrm{!}}} $
A few results that were given in my textbook were: $$ \begin{array}{l} {{S}_{0}\mathrm{{=}}{e}}\\ {{S}_{1}\mathrm{{=}}{e}}\\ {{S}_{2}\mathrm{{=}}{2}{e}}\\ {{S}_{3}\mathrm{{=}}{5}{e}}\\ {{S}_{4}\mathrm{{=}}{\mathrm{15}}{e}} \end{array} $$

The coefficients of $e$ piqued my interest, and so I used wolfram alpha to calculate $ {S}_{5} $, which came out to be equal to 52$e$. I looked up the sequence of coefficients of e on OEIS and it showed me a sequence of numbers known as the Bell numbers. I learned on Wikipedia that these numbers are used in Combinatorics, and give the maximum possible partitions of a set with given number of elements.

Anyhow, I attempted to solve the above series for $k$=2 and 3 to see if I could find a pattern linking bell numbers to the series. Here's what I did: $$ \begin{array}{l} {\mathop{\sum}\limits_{{n}\mathrm{{=}}{0}}\limits^{\mathrm{\infty}}{\frac{{n}^{2}}{n\mathrm{!}}}\mathrm{{=}}\mathop{\sum}\limits_{{n}\mathrm{{=}}{0}}\limits^{\mathrm{\infty}}{\frac{{n}{\mathrm{(}}{n}\mathrm{{-}}{1}{\mathrm{)}}\mathrm{{+}}{n}}{n\mathrm{!}}}\mathrm{{=}}\mathop{\sum}\limits_{{n}\mathrm{{=}}{0}}\limits^{\mathrm{\infty}}{\mathrm{(}\frac{1}{{\mathrm{(}}{n}\mathrm{{-}}{2}{\mathrm{)!}}}}\mathrm{{+}}\frac{1}{{\mathrm{(}}{n}\mathrm{{-}}{1}{\mathrm{)!}}}{\mathrm{)}}\mathrm{{=}}{e}\mathrm{{+}}{e}\mathrm{{=}}{2}{e}}\\ {\mathop{\sum}\limits_{{n}\mathrm{{=}}{0}}\limits^{\mathrm{\infty}}{\frac{{n}^{3}}{n\mathrm{!}}}\mathrm{{=}}\mathop{\sum}\limits_{{n}\mathrm{{=}}{0}}\limits^{\mathrm{\infty}}{\frac{{n}{\mathrm{(}}{n}\mathrm{{-}}{1}{\mathrm{)}}{\mathrm{(}}{n}\mathrm{{-}}{2}{\mathrm{)}}\mathrm{{+}}{3}{n}^{2}\mathrm{{-}}{2}{n}}{n\mathrm{!}}}\mathrm{{=}}\mathop{\sum}\limits_{{n}\mathrm{{=}}{0}}\limits^{\mathrm{\infty}}{\mathrm{(}\frac{1}{{\mathrm{(}}{n}\mathrm{{-}}{3}{\mathrm{)!}}}}\mathrm{{+}}{3}\frac{{n}^{2}}{n\mathrm{!}}\mathrm{{-}}{2}\frac{n}{n\mathrm{!}}{\mathrm{)}}\mathrm{{=}}{e}\mathrm{{+}}{3}{\mathrm{(}}{2}{e}{\mathrm{)}}\mathrm{{-}}{2}{\mathrm{(}}{e}{\mathrm{)}}\mathrm{{=}}{5}{e}} \end{array} $$ This method could be extended for any $k$, I believe, but will become tedious to calculate for larger $k$.
Needless to say, this didn't clear up any confusion for me. So could anyone please explain to me what's going on here? Any help regarding this will be much appreciated.


  • 1
    $\begingroup$ Reading the OEIS article I found this part: "Take the series 1^n/1! + 2^n/2! + 3^n/3! + 4^n/4! ... If n=1 then the result will be e, about 2.71828. If n=2, the result will be 2e. If n=3, the result will be 5e. This continues, following the pattern of the Bell numbers: e, 2e, 5e, 15e, 52e, 203e, etc. - Jonathan R. Love (japanada11(AT)yahoo.ca), Feb 22 2007" You thus might want to contact the given email address for help or a proof, as you are not the first one to discover it. $\endgroup$
    – Dirk
    Commented Apr 7, 2017 at 10:57
  • $\begingroup$ en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind ... You will find the Stirling numbers of the second time are an interesting grading of the Bell Numbers. ... The numerator of your last equation $n(n-1)(n-2)+3n(n-1)+n$ might work better ? $\endgroup$ Commented Apr 7, 2017 at 10:59
  • 1
    $\begingroup$ That's known as Dobinsky Formula $\endgroup$
    – G Cab
    Commented Dec 31, 2017 at 19:03

2 Answers 2


$\newcommand\D{\text{D}}$ $\newcommand\Stir[2]{ {#1 \brace #2} }$ $\newcommand\diff[2]{\frac{\text{d} #1}{\text{d} #2}}$It is well known that stirling numbers of the second kind $\smash{\Stir{a}{b}}$ are related to the operator $\smash{x\D\equiv x\diff{}{x}}$


Which can be confirmed by checking that the coefficients of $x^{k-j}\D^j$ obey the recurrence relation for Stirling numbers of the second kind.

Then operating $\eqref{1}$ on $e^x$ we have, since $\smash{\D^j(e^x)}=e^x$

$$(x\D)^ke^x= e^x\sum_{j=0}^{k}\Stir{k}{j}x^{k-j}\tag{2}\label{2}$$

by writing $\smash{e^x=\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}}$the left hand side of $\eqref{2}$ is




so putting $x=1$ in $\eqref{3}$ gives


then because the $n^{\text{th}}$ Bell number $B_n$ is given by


we have your relation by substituting $\eqref{5}$ in to $\eqref{4}$:

$$ \sum_{n=0}^{\infty}\frac{n^k}{n!}=eB_n\tag{6}\label{6}$$


A good way to start on $$S_{k}=\sum\limits_{n=0}^{\infty}\frac{{n}^{k}}{n!} $$ is to express $S_k$ in terms of the previous $S_j$ using the binomial theorem.

If $k \ge 1$,

$\begin{array}\\ S_{k} &=\sum\limits_{n=0}^{\infty}\dfrac{{n}^{k}}{n!}\\ &=\sum\limits_{n=1}^{\infty}\dfrac{{n}^{k}}{n!}\\ &=\sum\limits_{n=1}^{\infty}\dfrac{{n}^{k-1}}{(n-1)!}\\ &=\sum\limits_{n=0}^{\infty}\dfrac{(n+1)^{k-1}}{(n!}\\ &=\sum\limits_{n=0}^{\infty}\dfrac1{n!}(n+1)^{k-1}\\ &=\sum\limits_{n=0}^{\infty}\dfrac1{n!}\sum_{j=0}^{k-1}\binom{k-1}{j}n^j\\ &=\sum_{j=0}^{k-1}\binom{k-1}{j}\sum\limits_{n=0}^{\infty}\dfrac1{n!}n^j\\ &=\sum_{j=0}^{k-1}\binom{k-1}{j}S_j\\ \end{array} $

(As many of my answers, nothing here is original.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .