How to show $\frac 1 e \sum\limits_{n=0}^{+\infty}\frac{n^x}{n!}$ is always an integer for $x \in \mathbb N$

Question

The series $$\frac 1 e \sum_{n=0}^{+\infty}\frac{n^x}{n!}$$ converges to an integer by some of my computations using wolfram alpha with $x \in \mathbb{N}$. I don't succed to show that this series is always integer by evaluating using the partial sum of it because the partial sum is complicated for every integer $x \geq 0$ .Then I have this question:

What is the partial sum of :$$\frac 1 e \sum_{n=0}^{+\infty}\frac{n^x}{n!}$$
for fixed integer $x$ such that $0\leq x < n$ and how I can prove it always integer ?

Answer 1

$\def\e{\mathrm{e}}$Denote$$ S_m = \frac{1}{\e} \sum_{k = 0}^\infty \frac{1}{k!} \prod_{j = 0}^{m - 1} (k - j), $$ then$$ S_m = \frac{1}{\e} \sum_{k = m}^\infty \frac{1}{k!} \prod_{j = 0}^{m - 1} (k - j) = \frac{1}{\e} \sum_{k = m}^\infty \frac{1}{(k - m)!} = \frac{1}{\e} \sum_{k = 0}^\infty \frac{1}{k!} = 1 \in \mathbb{Z}. $$

Now denote$$ T_m = \frac{1}{\e} \sum_{k = 0}^\infty \frac{k^m}{k!}. $$ Note that for any $m \in \mathbb{N}$, $T_m$ is a linear combination of $T_0, \cdots, T_{m - 1}, S_m$ with combination coefficients being integers, by induction on $m$, $T_m \in \mathbb{Z}$.

Answer 2

For any integer $x \ge 0$, let: $$T(x) = \sum_{ n \ge 0 } \frac{n^x}{n!}$$ Then you have $T(0) = \sum_\limits{ n \ge 0 } \frac{1}{n!} = e $, and for any integer $x\ge0$,

$$ T(x+1) = \sum_\limits{ n \ge 0 } \frac{n^{x+1}}{n!} = \sum_\limits{ n \ge 0 } \frac{ n \cdot n^x}{n!} = \sum_\limits{ n \ge 1 } \frac{n^x}{(n-1)!} = \sum_\limits{ n \ge 0 } \frac{(n+1)^x}{n!} $$

Now you can expand the inner term using the binomial theorem to get a recurrence formula: $$ T(x+1) = \sum_\limits{ n \ge 0 }\sum_\limits{ 0 \le k \le x } \frac{C_k^x \cdot n^k}{n!} =\sum_\limits{ 0 \le k \le x } C_k^x \sum_\limits{ n \ge 0 } \frac{ n^k}{n!} = \sum_\limits{ 0 \le k \le x } C_k^x \cdot T(k) $$

Therefore, you can easily show that $T(x)$ is an integer multiple of $T(0) = e$ for any $x \in \mathbb{N} $.

Answer 3

You are dealing with Touchard polynomials. We have $$ \sum_{n\geq 0}\frac{e^{nz}}{n!} = \exp\left(e^z\right)$$ and by applying $\frac{d^m}{dz^m}$ to both sides we get $$ \sum_{n\geq 0}\frac{n^m e^{nz}}{n!} = Q_m(e^z)\cdot\exp\left(e^z\right),\qquad Q_m(x)\in\mathbb{Z}[x] $$ so the evaluation at $z=0$ leads to $$ \frac{1}{e}\sum_{n\geq 0}\frac{n^m}{n!} = Q_m(1)\in\mathbb{Z} $$ as wanted.