Calculate $\sum_{n=1}^\infty\frac{n^x}{n!}$

Question

I want to evaluate function defined by following sum: $$\sum_{n=1}^\infty\frac{n^x}{n!}$$ I was thinking about writing Taylor series expansion for it. However my try resulted in sum that looks even harder to calculate: $$\sum_{n=1}^{\infty}\frac{\ln^k(n)}{n!}$$Thanks for all the help in solving this problem.

Answer 1

For $x=0$, you recognize

$$\sum_{n=1}^\infty\frac1{n!}=e-1.$$

For $x=1$,

$$\sum_{n=1}^\infty\frac n{n!}=\sum_{n=1}^\infty\frac1{(n-1)!}=e.$$

For $x=2$,

$$\sum_{n=1}^\infty\frac{n^2}{n!}=\sum_{n=1}^\infty\frac{n(n-1)+n}{n!}=\sum_{n=2}^\infty\frac1{(n-2)!}+\sum_{n=1}^\infty\frac1{(n-1)!}=2e.$$

For larger powers $x$, the sum will depend on the decomposition of $n^x$ as a sum of falling factorials $(n)_x$, which is given by the Stirling numbers of the second kind, https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind#Definition. Hence the sums are given by $e$ times the sum of the first Stirling numbers, which are the Bell numbers.

There does not seem to be an easy generalization to negative nor fractional powers. Asymptotic expressions for the Bell numbers are available.

Answer 2

The sum is the $x^{\text{th}}$ Bell number times $e$. The recursive sequence of Bell's number is given by $$B_{n+1}=\sum_{k=0}^{n} \binom{n}{k} B_k$$

Answer 3

Proof of the Dobiniski's formula:

See, $$\sum_{n=0}^{\infty} \frac{n^k}{n!}=\underbrace{\frac{d}{dx}(x\frac{d}{dx}(x.....(\frac{d}{dx}e^x)))),}_\text{$k$ times}$$ at $x=1$.

$$=\underbrace{\frac{d}{dx}(x\frac{d}{dx}(......\frac{d}{dx}(x^2e^x+xe^x))),}_\text{$k-2$ times}$$ at $x=1$.....(1)

Also , $eB_k=\frac{d^k}{dt^k}(e^{e^t})$ at $t=0$

$=\frac{d^{k-2}}{dt^{k-2}}((e^t)^2e^{e^t}+e^te^{e^t})$ at $t=0$ ....(2)

See, (1) and (2) have exact the same structure. Also $e^t=1, e^{e^t}=e$ at $t=0$. And, at $x=1, e^x=e$.

Hence, we get $$eB_k=\sum_{n=0}^{\infty} \frac{n^k}{n!}$$.

Answer 4

We can rewrite the sum as $$ \eqalign{ & F(x) = \sum\limits_{1\, \le \,n} {{{n^{\,x} } \over {n!}}} = \sum\limits_{0\, \le \,n} {{{\left( {n + 1} \right)^{\,x} } \over {\left( {n + 1} \right)!}}} = \cr & = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,j} {\left( \matrix{ x \cr j \cr} \right){{n^{\,j} } \over {\left( {n + 1} \right)!}}} } = \cr & = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,j} {{{x^{\,\underline {\,j\,} } } \over {j!}}{{n^{\,j} } \over {\left( {n + 1} \right)!}}} } = \cr & = \sum\limits_{0\, \le \,j} {\left( {\sum\limits_{0\, \le \,n} {{{n^{\,j} } \over {\left( {n + 1} \right)!}}} } \right)} {{x^{\,\underline {\,j\,} } } \over {j!}} \cr} $$

which gives the expansion of $F(x)$ as a Newton series, the inner sum being "near-Bell" numbers

If we replace the falling Factorial with $$ x^{\,\underline {\,j\,} } = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,j} \right)} {\left( { - 1} \right)^{\,j - k} \left[ \matrix{ j \cr k \cr} \right]x^{\,k} } $$ we unfortunately get an alternating diverging series for the coefficients of $x^k$.

However the actual Bell numbers intevene into $$ \eqalign{ & F(x + 1) = \sum\limits_{0\, \le \,n} {{{\left( {n + 1} \right)^{\,x} } \over {n!}}} = \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{0\, \le \,n} {{{n^{\,k} } \over {n!}}} } \right)\left( \matrix{ x \cr k \cr} \right)} = \cr & = e\sum\limits_{0\, \le \,k} {B_{\,k} \left( \matrix{ x \cr k \cr} \right)} = e\sum\limits_{0\, \le \,k} {{{B_{\,k} } \over {k!}}x^{\,\underline {\,k\,} } } \cr} $$ wherefrom we also get the interesting recurrence $$ \eqalign{ & F(x + 1) = \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{0\, \le \,n} {{{n^{\,k} } \over {n!}}} } \right)\left( \matrix{ x \cr k \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k} {\left( {\left[ {k = 0} \right] + \sum\limits_{1\, \le \,n} {{{n^{\,k} } \over {n!}}} } \right)\left( \matrix{ x \cr k \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k} {\left( {\left[ {k = 0} \right] + F(k)} \right)\left( \matrix{ x \cr k \cr} \right)} = \cr & = 1 + \sum\limits_{0\, \le \,k} {\left( \matrix{ x \cr k \cr} \right)F(k)} \cr} $$