Limit $c^n n!/n^n$ as $n$ goes to infinity 
Let $c>0$ be a real number. I would like to study the convergence of $a_n := c^n n!/n^n$, when $n \to \infty$, in function of $c$.

I know (from this question) that $n!>(n/e)^n$, so that $c^n n!/n^n>1$ for $c ≥e$. But this doesn't imply that the sequence goes to infinity. And I'm not sure what to do for $c<e$. I tried usual tests (D'Alembert...), without any success. I would like to avoid using Stirling approximation.
Thank you for your help!
 A: If we prove that the sequence $ a_n = \left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$ is decreasing towards $e$, from
$$ n=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right),\qquad n!=\frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k }=\frac{n^{n+\frac{1}{2}}}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{k+\frac{1}{2}} }\tag{1} $$
we get:
$$ \frac{n!}{n^n}\leq \frac{\sqrt{n}}{e^{n-1}} \tag{2} $$
and by exploiting the fact that the sequence $b_n=\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}-\frac{1}{12 n}}$ is increasing towards $e$ and the infinite product $\prod_{k\geq 1}\left(1+\frac{1}{k}\right)^{-\frac{1}{12k}}$ is converging to a constant $0<C<1$ we get
$$\boxed{\; C\frac{\sqrt{n}}{e^{n-1}}\leq \frac{n!}{n^n}\leq \frac{\sqrt{n}}{e^{n-1}}\;}\tag{3} $$
that is a weak form of Stirling's inequality, elementary but powerful enough for our purposes.
The properties of the involved sequences follow from the fact that in a neighbourhood of the origin we have
$$ \frac{1}{\log(1+x)} = \frac{1}{x}+\frac{1}{2}-\frac{1}{12}x+\ldots = \frac{1}{x}+\sum_{n\geq 1}G_n x^{n-1}\tag{4} $$
and the Gregory coefficients $G_n$ behave like $\frac{1}{n\log^2 n}$ in absolute value and have alternating signs.
In any case, there is a rather short way for proving Stirling's inequality through creative telescoping (yes, I am very proud of the linked answer).
A: You have (Stirling Formula)
$$n!\sim \sqrt{2\pi n} \frac {n^n}{e^n}.$$
So
$$\frac {c^n n!}{n^n}\sim \frac{c^n \sqrt{2\pi n} n^n}{e^n n^n}=\left(\frac ce\right)^n\sqrt{2\pi n}.$$
Therefore, if $c/e<1$ you get the limit $0$, and if $c\geq 1$, you get $+\infty$.
It all depends whether or not $c<e$.
A: Since you do not want Stirling approximation, just use the classical ratio test
$$a_n=\frac {c^n n!}{n^n}\implies \frac{a_{n+1}}{a_n}=c \left(\frac{n}{n+1}\right)^n=\frac c {\left(1+\frac 1 n\right)^n}$$ which tends to $\frac c e$ when $n\to \infty$.
A: Hint:
Taking the logarithm, you want to discuss the convergence of 
$$l_n=\sum_{k=1}^n\log k+n\log c-n\log n=\sum_{k=1}^n\log\frac kn+n\log c.$$
Seeing the sum as a Riemanian integral, you get a first approximation
$$l_n\approx n\int_{x=1/n}^1\log x\,dx+n\log c=-n+\log n+1+n\log c$$ which probably allows you to conclude for $c\ne e$ as the linear terms dominate.
For a more accurate result (when $c=e$), you should use the Euler-Maclaurin summation formula.
