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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!

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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 $n\to \infty$.

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

$$ \frac{1}{\log(1+x)} = \frac{1}{x}+\frac{1}{2}-\frac{1}{12x}+\ldots = \frac{1}{x}+\sum_{n\geq 1}G_n x^{n-1}\tag{4} $$

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).

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 $c<e$.

**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.

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