For fixed $n\geq 1$, consider the sum $$S_n:=\sum_{k=1}^\infty \frac{k^n}{e^k}.$$ If you compute this sum for some small values of $n$, you will see that it is remarkably close to $n!$. The fact that it is somehat close to $n!$ is not too surprising because $S_n$ "should" be similar to the following integral $$n! = \int_0^\infty \frac{x^n}{e^x}dx.$$ However, the precision of the approximation seems to be surprisingly good (check some small values on Wolfram Alpha and see for yourself!).
Question: Is there a good way to show that $S_n$ is bounded above by $(1+\varepsilon) n!$ for every $n\geq1$ where $\varepsilon$ is some explicit positive constant which is "close" to $0$?
After checking some small values of $n$, it appears that $n=3$ might be the worst value for an upper bound of this type. Note that, for many values of $n$ (perhaps for most $n$), it is actually the case that $S_n<n!$.
If you think of $S_n$ as a Riemann sum approximating the integral, then the $k$th summand is approximating the area under the curve between $x=k-1$ and $x=k$ by the value at the right endpoint of this interval (namely, by $\frac{k^n}{e^k}$). The function $\frac{x^n}{e^x}$ is increasing for $0\leq x<n$ and decreasing for $x>n$. Therefore, for $k\leq n$, the $k$th summand is overestimating the area and for $k\geq n+1$ it is underestimating it. Magically, it seems that the overestimating terms and the underestimating terms cancel out beautifully and give us a result which is really close to $n!$. Does anyone know whether this phenomenon occurs for all $n$ and whether there is a good reason why this would be occur?
By the way, there is a connection with Eulerian numbers here, if anyone is interested. It turns out that $$S_n = \frac{e\cdot \sum_{m=0}^{n-1}A(n,m)e^m}{(e-1)^{n+1}}$$ where $A(n,m)$ is the number of permutations of $1,\dots,n$ with exactly $m$ "ascents." Interestingly, $A(n,0)+A(n,1)+\cdots +A(n,n-1)=n!$ (since every permutation has some number of ascents) but this doesn't seem to be terribly helpful since there are these factors of $e$ and $e-1$ floating around everywhere.
For any $\alpha > 0$ we have that $\sum_{k\geq 1}e^{-\alpha k}=\frac{1}{e^{\alpha}-1}$, hence by differentiating both sides $n$ times with respect to $\alpha$ we get that $$ (-1)^n\sum_{k\geq 1} k^n e^{-\alpha k} = \frac{d^n}{d\alpha^n}\,\frac{1}{e^\alpha-1}=\frac{d^n}{d\alpha^n}\left[\frac{1}{\alpha}+\sum_{m\geq 1}\frac{B_m}{m!}\alpha^{m-1}\right] \tag{1}$$ hence: $$ (-1)^n \sum_{k\geq 1} k^n e^{-\alpha k} = \frac{n!(-1)^n}{\alpha^{n+1}}+\sum_{m\geq n+1}\frac{B_m}{m!}\cdot\frac{(m-1)!}{(m-n-1)!}\alpha^{m-n-1}\tag{2} $$ and by evaluating at $\alpha=1$ it comes the magic: $$ \sum_{k\geq 1}\frac{k^n}{e^k} = \color{red}{n!}+(-1)^n \sum_{m\geq n+1}\frac{B_m}{m(m-n-1)!}\tag{3} $$ The behaviour of Bernoulli numbers (I am going to talk about Bernoulli numbers with even index, since every Bernoulli number with odd index is zero, with the exception of $B_1$) is quite erratic: till $B_{12}$ they are all less than one in absolute value, then their absolute value starts growing pretty fast: $|B_{2 n}| \sim 4 \sqrt{\pi n} \left(\frac{n}{ \pi e} \right)^{2n}$ for large values of $n$. Since the remainder series in $(3)$ converges pretty fast and the first Bernoulli numbers are essentially negligible, for small values of $n$ (namely $n\leq 16$) $S_n$ and $n!$ are very close, as conjectured.