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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.

Thanks

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    $\begingroup$ 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
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    $\begingroup$ That's known as Dobinsky Formula $\endgroup$
    – G Cab
    Commented Dec 31, 2017 at 19:03

2 Answers 2

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

$$(x\D)^k\equiv\sum_{j=0}^{k}\Stir{k}{j}x^{k-j}\D^j\tag{1}\label{1}$$

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

$$(x\D)^ke^x=\sum_{n=0}^{\infty}\frac{n^k}{n!}x^n$$

therefore

$$\sum_{n=0}^{\infty}\frac{n^k}{n!}x^n=e^x\sum_{j=0}^{k}\Stir{k}{j}x^{k-j}\tag{3}\label{3}$$

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

$$\sum_{n=0}^{\infty}\frac{n^k}{n!}=e\sum_{j=0}^{k}\Stir{k}{j}\tag{4}\label{4}$$

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

$$B_n=\sum_{j=0}^{k}\Stir{k}{j}\tag{5}\label{5}$$

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

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

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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.)

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