Prove that the series $\sum_{n=0}^\infty \frac{(n!)^n}{e^{n!}}$ does not converge.

Question

Prove that the series diverges. $$\sum_{n=0}^\infty \frac{(n!)^n}{e^{n!}}$$

IT'S SUPPOSED TO BE e^(n!) in the denominator if it isn't clear.

I am having a lot of difficulties with this problem. I want to solve it with the comparison test or ratio test. Any tips?

Thank You

Answer 1

The series $\displaystyle\sum_{n=0}^\infty \frac{(n!)^n}{e^{n!}}$ is convergent by the Root test: we have that $$\lim_{n\to \infty}\frac{n!}{e^{(n-1)!}}\leq \lim_{n\to \infty}\frac{n!}{1+(n-1)!+\frac{1}{2}((n-1)!)^2}=\lim_{n\to \infty}\frac{1}{\frac{1}{n!}+\frac{1}{n}+\frac{(n-1)!}{2n}}=0$$ where we used the inequality $e^x\geq 1+x+\frac{x^2}{2}$ for $x\geq 0$.

P.S. Your inequality $\frac{(n!)^n}{e^{n!}} > \frac{(n!)^n}{(e^n)^n}$ is wrong because $e^{n!}>e^{n^2}=(e^n)^n$ for $n\geq 4$.

Answer 2

$$n\log n!-n!\sim n^2\log n-n!<-(n-3)!$$ decreases pretty fast. The series converges.

Answer 3

The series converges by root test

$$\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty\frac{(n!)^n}{e^{n!}}\implies \sqrt[n] {a_n}=\frac{n!}{e^{(n-1)!}}\to 0$$

indeed by squeeze theorem

$$0\le\frac{n!}{e^{(n-1)!}}\le\frac{(n-1!)^2}{e^{(n-1)!}}\to 0$$

Answer 4

For $n \in \mathbb{N}$, we have : $ e^{n!} = (((e^1)^2)^{...})^n $

Hence : $ \frac{(n!)^n}{e^{n!}} = \left(\frac{n!}{((e^1)^{...})^{n-1}}\right)^n \le \frac{1}{n^n} $

So this is less than the $\frac{1}{n^2}$ serie which converges, the serie must diverge