Value of $\sum_{n=0}^{\infty}\frac{n^k}{n!}$ for $k \in \mathbb{N}$

Question

I'm trying to find an explicit formula for the aforementioned sum (I am not sure that it is possible).

The first values are as follows, with $S_{k}=\sum_{n=0}^{\infty}\frac{n^k}{n!}$ :

$S_0=e$, $S_1=e$, $S_2=2e$, $S_3=5e$, $S_4=15e$, $S_5=52e$.

I couldn't see any obvious pattern from this, but I was able to find a recurrence relation for $S_k$ , by noticing that

$S_{k+2}=\sum_{n=1}^{\infty}\frac{n^{k}\left ( n-1+1 \right )}{\left ( n-1 \right )!}$ $=S_{k+1}+\sum_{n=1}^{\infty}\frac{\left ( n+1 \right )^{k}}{\left ( n-1 \right )!}$.

By expanding $(n+1)^{k}$, we get $S_{k+2}=S_{k+1}+\sum_{i=0}^{k}\binom{k}{i}S_{i+1}$ .

Problem is, I am unable to solve this recurrence relation.

It kind of reminds me of the relation between Bernoulli's numbers : $\sum _{k=0}^{n}{n+1 \choose {k}}B_{k}=0$. It is possible to find an explicit formula for Bernoulli's numbers, but it seems rather complicated to me (see this thread Explicit formula for Bernoulli numbers by using only the recurrence relation )

Perhaps it is possible to use a similar method to find an explicit formula for $S_k$ ?

EDIT : Thanks for the help. We can take the derivative inside the series since f is a power series with an infinite radius of convergence.

f is the exponential function so $f'=f$ :

$xf'\left ( x \right )=xf\left ( x \right )$,

$x\frac{\mathrm{d}}{\mathrm{d} x}\left ( xf'\left ( x \right ) \right )=\left ( x^{2}+x \right )f\left ( x \right )$,

$x\frac{\mathrm{d}}{\mathrm{d} x}\left (\left ( x^{2}+x \right )f\left ( x \right ) \right )=\left ( x^{3}+3x^{2}+x \right )f\left ( x \right )$

I can't seem to find a general formula for $\left ( x\frac{\mathrm{d} }{\mathrm{d} x} \right )^{k}f\left ( x \right )$ though. I'm pretty sure it involves binomial coeffficients...

If we have one, I see that by plugging $x=1$ we get a formula for $S_k$

You might find this helpful: https://en.wikipedia.org/wiki/Bell_number — carmichael561, Apr 12 '18 at 04:53
@charmichael561 thanks a lot. So $S_k=eB_k$... I'm sad to learn that we don't have any explicit formula for Bell numbers :( — Skywear, Apr 12 '18 at 05:34
@AlexD Funny indeed !
That's just a question that I asked myself. I started calculating $\sum_{n=0}^{\infty}\frac{n}{n!}$, $\sum_{n=0}^{\infty}\frac{n^{2}}{n!}$ for no specific reason, and then wondered if I could find a more general formula. What about you ? — Skywear, Apr 12 '18 at 05:38
@carmichael561 my bad, there is one https://imgur.com/a/TadCy — Skywear, Apr 12 '18 at 06:13
Also: https://en.wikipedia.org/wiki/Touchard_polynomials (evaluated at $x=1$) — Jack D'Aurizio, Apr 12 '18 at 10:32
I agree with these comments, you can at best define the integers in your coefficients and establish some recurrence equations or other properties. These numbers are deeply studied and there's no reason to be sad because they have no known closed form (the set of sequences which can be written in closed form is extremely sparse anyway) — Tal-Botvinnik, Apr 12 '18 at 14:11
See also: Compute $\frac{1}{e}\sum\limits_{n=0}^{\infty}\frac{n^{k}}{n!}$ for $k=0, 1, 2 ... $ and Evaluating sum $\sum \limits_{n=0}^\infty\frac{n^k}{n!}$. — Martin Sleziak, Nov 02 '19 at 06:19

score 2 · Answer 1 · answered Apr 12 '18 at 04:19

Let $$f(x)=\sum_{n\geq 0} \frac{x^n}{n!}$$ And note that

$$xf'(x) =\sum_{n\geq 0} \frac{nx^n}{n!}$$

Repeat $k$ times and you'll have

$$(x\frac{d}{dx})^k f(x)=\sum_{n\geq 0} \frac{n^kx^n}{n!}.$$

Now, what is $f(x)$? And $f(1)$? Also, can you justify taking the derivatives inside the series?

Value of $\sum_{n=0}^{\infty}\frac{n^k}{n!}$ for $k \in \mathbb{N}$

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