Sum of the series $\sum_{k=0}^\infty k^ne^{-k}$ for a positive integer $n$

Question

How can we calculate the sum of the series $\sum_{k=0}^\infty k^ne^{-k}$ for a positive integer $n$? I tried: $$\sum_{k=0}^\infty k^ne^{-k} =(-1)^n \sum_{k=0}^\infty \frac{d^n}{dy^n}\bigg|_{y=1}e^{-ky}=(-1)^n \frac{d^n}{dy^n}\bigg|_{y=1}\sum_{k=0}^\infty e^{-ky} = (-1)^n\frac{d^n}{dy^n}\bigg|_{y=1} \frac{1}{1-e^{-y}}$$ but I got stuck for getting $\frac{d^n}{dy^n}\bigg|_{y=1} \frac{1}{1-e^{-y}}$ for general $n$. Is there a way for getting this, or is there another approach?

P.S. What I am actually wanting to compute is the value $\lim_{n\to \infty} (\sum_k k^n e^{-k})/n!$.

From @Will Jagy's answer I've got the following: the function $f(x)=x^ne^{-x}$ is increasing for $x<n$ and decreasing for $x>n$, so $\sum_{k=0}^\infty k^ne^{-k}= \sum_{k=1}^{n-1} k^ne^{-k} + n^ne^{-n}+\sum_{k=n+1}^\infty k^ne^{-k} \leq \int_0^n x^ne^{-x}dx+n^ne^{-n}+\int_n^\infty x^ne^{-x}dx=n!+n^ne^{-n}$. But doesn't $n^ne^{-n}$ grow too fast as $n\to \infty$?

Answer 1

In general $$\sum_{k \geq 0} k^n e^{-k} = \sum_{l = 1}^n \left\langle \begin{matrix}n\\l\end{matrix}\right\rangle\frac{e^l}{(e-1)^{n+1}}, $$ where $$\left\langle \begin{matrix}n\\l\end{matrix}\right\rangle := \sum_{j = 0}^l (-1)^j {{n + 1} \choose j} (l + 1 - j)^n$$ are the Eulerian numbers, sometimes denoted $T(n, l)$, respectively. (We can also express the Eulerian numbers in terms of the Stirling numbers of the second kind.) See Richard R. Forberg's comment from 2015 February 15 in OEIS A008292 (Triangle of Eulerian numbers $T(n, k)$ ($n \geq 1$, $1 \leq k \leq n$) read by rows), and see the rest of the entry for many beautiful combinatorial interpretations of the Eulerian Numbers. For small $n$ we have

$$\begin{array}{c|cc} n & \sum_{k \geq 0} k^n e^{-k} \\ \hline 0 & \displaystyle \frac{1}{e - 1} & 0.58197\ldots\\ 1 & \displaystyle \frac{e}{(e - 1)^2} & 0.92067\ldots \\ 2 & \displaystyle \frac{e (e + 1)}{(e - 1)^3} & 1.99229\ldots \\ 3 & \displaystyle \frac{e (e^2 + 4 e + 1)}{(e - 1)^4} & 6.00651\ldots \\ 4 & \displaystyle \frac{e (e^3 + 11 e^2 + 11 e + 1)}{(e - 1)^4} & 24.00333\ldots \\ \end{array}$$

More generally (i.e., for arbitrary $n$) the definition of the polylogarithm gives $$\sum_{k \geq 0} k^n e^{-k} = \operatorname{Li}_{-n}\left(\frac{1}{e}\right) .$$

We can of course discard the $k = 0$ term in our sums, in which case taking $n = -1$ gives the special value $$\sum_{k \geq 1} \frac{e^{-k}}{k} = \operatorname{Li}_1\left(\frac{1}{e}\right) = 1 - \log(e - 1) ) = 0.45867 \ldots .$$

For large $n$ we should expect that the sum is asymptotic to $\int_0^{\infty} x^n e^{-x} \,dx = n!$ (already the numerical approximations in our table for small $n$ suggest as much), and hence that $$\lim_{n \to \infty} \frac{\sum_{k \geq 0} k^n e^{-k}}{n!} = 1 .$$ Indeed, the answers to this question (h/t Gary) yield the explicit asymptotic $$\frac{\sum_{k \geq 0} k^n e^{-k}}{n!} - 1 \in O((2 \pi)^{-n}) .$$

Answer 2

The series $\sum_{k=0}^\infty k^ne^{-k}$ is approximated well enough by $$ I_n = \int_0^\infty x^n e^{-x} dx $$ At the same time, $$ I_n = n!$$ I did that by induction and integration by parts.

So the limit you care about is $1$

=========================================================

Let's see, $x^n e^{-x}$ is increasing until its maximum at $x = n,$ then decreases down to zero at infinity.

For the estimates, for monotonic functions we have:

if we have $f(x) > 0$ and $f'(x) > 0,$ then $$ \int_{a-1}^{b} \; f(x) \; dx \; < \; \sum_{j=a}^b \; f(j) \; < \; \int_{a}^{b+1} \; f(x) \; dx $$

if we have $f(x) > 0$ but $f'(x) < 0,$ then $$ \int_a^{b+1} \; f(x) \; dx \; < \; \sum_{k=a}^b \; f(k) \; < \; \int_{a-1}^b \; f(x) \; dx $$