How good is this approximation of $n$ factorial? 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. 
 A: By the Euler-Maclaurin formula,
$$S_n\approx n!-\frac1{2e}$$
You can get better approximations by taking more terms.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\left.\sum_{k = 1}^{\infty}{k^{n} \over \expo{k}}\right\vert_{\ n\ \geq\ 0} & =
-\delta_{n0} + \sum_{k = 0}^{\infty}{k^{n} \over \expo{k}} = 
-\delta_{n0} + \int_{0}^{\infty}{k^{n} \over \expo{k}}\,\dd k +
\left.{1 \over 2}{k^{n} \over \expo{k}}\right\vert_{\ k\ =\ 0} -2
\int_{0}^{\infty}{\Im\pars{\bracks{\ic x}^{n}/\expo{\ic x}} \over
\expo{2\pi x} - 1}\,\dd x
\\[5mm] & = -\,{1 \over 2}\,\delta_{n0} + n! -
2\left\{\begin{array}{lcl}
\ds{-\pars{-1}^{n/2}\int_{0}^{\infty}{x^{n}\sin\pars{x}
\over \expo{2\pi x} - 1}\,\dd x} & \mbox{if} & \ds{n}\ \mbox{is}\ even
\\[3mm]
\ds{\phantom{-\,}\pars{-1}^{\pars{n - 1}/2}\int_{0}^{\infty}{x^{n}\cos\pars{x}
\over \expo{2\pi x} - 1}\,\dd x} & \mbox{if} & \ds{n}\ \mbox{is}\ odd
\end{array}\right.
\end{align}

$$
\left.\vphantom{\large A}n!\,\right\vert_{\ n\ \geq\ 0} =
\sum_{k = 1}^{\infty}{k^{n} \over \expo{k}} +
{1 \over 2}\,\delta_{n0} +
2\left\{\begin{array}{lcl}
\ds{-\pars{-1}^{n/2}\int_{0}^{\infty}{x^{n}\sin\pars{x}
\over \expo{2\pi x} - 1}\,\dd x} & \mbox{if} & \ds{n}\ \mbox{is}\ even
\\[3mm]
\ds{\phantom{-\,}\pars{-1}^{\pars{n - 1}/2}\int_{0}^{\infty}{x^{n}\cos\pars{x}
\over \expo{2\pi x} - 1}\,\dd x} & \mbox{if} & \ds{n}\ \mbox{is}\ odd
\end{array}\right.
$$


Relative Error plot:



Absolute Error plot: It's worsening as $\ds{n > 16}$. Such behaviour is fully discussed in
  @Jack D'Aurizio answer.


A: 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.
