How fast does $\frac{1}{n!} \sum_{k=0}^n k(n-k)!$ grow, as a function of $n$? Consider the sum $$\frac{1}{n!} \sum_{k=0}^n k(n-k)!$$ How fast does it grow (as a function of $n$)? I can prove that it grows slower than or equal to linearly (but I expect this is very crude), so I am looking for a better bound of growth. 
The closed form I found in Wolfram Alpha includes subtracting some factorials, so is not a keen indicator of growth. I tried bounding this using the Gamma function: It is $\le \int_0^n x\Gamma(n-x+1)$, but I am not sure what I can do from here.
Where does this crop up? Well, it is the expected stopping time to a common "puzzle" question.
 A: We can rewrite the series as 
$$\sum_{k = 0}^n k(n-k)! = \sum_{k = 0}^n (n-k)k! = n\sum_{k = 0}^n k! - \sum_{k=0}^n k \cdot k! = n\sum_{k = 0}^n k! - (n+1)!+1$$
where the last equality follows from the identity $\sum_{i = 0}^n k \cdot k! = (n+1)! - 1$. If this post is to be believed, we have an asymptotic expansion
$$ \sum_{k = 0}^n k! = n!\left(1 + \frac{1}{n} + \frac{1}{n^2} + O\left(\frac{1}{n^3}\right)\right)$$
Using these two observations to your series gives
$$\frac{1}{n!}\sum_{k=0}^nk(n-k)! = n\left(1 + \frac{1}{n} + \frac{1}{n^2} + O\left(\frac{1}{n^3}\right)\right) - (n+1)+\frac{1}{n!} = \frac{1}{n} + O\left(\frac{1}{n^2}\right)$$
And hence as $n \to \infty$, the series goes to $0$.
(Disclaimer: I don't really know much about asymptotic series, so some of my reasoning there may be off).
A: This sum is bounded above, and it's easy to show a crude estimate.
Let
$$f_n(x) = \sum_{k=0}^n\,\frac{(n-k)!}{n!}\,x^k=\sum_{k=0}^n\,\frac{1}{\binom{n}k}\,\frac{x^k}{k!}.$$
Then
\begin{align}
f_n'(x) = \sum_{k=0}^n\,\frac{k\,(n-k)!}{n!}\,x^k
=\sum_{k=1}^n\,\frac{1}{\binom{n}k}\,\frac{x^{k-1}}{(k-1)!}
&\leq
\sum_{k=1}^n\,\frac{x^{k-1}}{(k-1)!}\\
&=
\sum_{k=0}^{n-1}\,\frac{x^{k}}{k!}\\
&\leq
\sum_{k=0}^{\infty}\,\frac{x^{k}}{k!}=\exp(x),
\end{align}
and observe that this bound does not depend on $n$.
Moreover, notice that $\frac{1}{n!} \sum_{k=0}^n k(n-k)! =$ $f'(1)$.
It follows that
$$\frac{1}{n!} \sum_{k=0}^n k(n-k)! \leq e$$
A: $$\sum_{k=0}^{n}(n-k)k! = \int_{0}^{+\infty}\sum_{k=0}^{n}(n-k)x^k e^{-x}\,dx =\int_{0}^{+\infty}\underbrace{\frac{n-(n+1)x+x^{n+1}}{(1-x)^2}}_{\ll x^{n-1}}e^{-x}\,dx$$
hence the LHS is $\ll (n-1)!$ and the wanted limit is clearly zero.
