Divergent infinite series $n!e^n/n^n$ - simpler proof of divergence?

Question

$$ \sum_{n=1}^\infty \frac{n!e^n}{n^n} $$ Where $e$ is Euler's number.

Recently on calculus class we were covering convergence tests and my group got stuck with this infinite series. Our calculus teacher told us that it diverges but he doesn't see any elementary way or simple test to prove it. He told us something about Euler's Gamma function and said that he knows it diverges because of Stirling's formula but this is not quite our level yet.

I tried to think about it myself but I established only that this is some kind of an edge case. We could generalize the problem to series of $\frac{n!a^n}{n^n}$. Then, from the ratio test we know that it converges for $a <e$ and diverges for $a>e$. For $a=e$ the test is inconclusive. I've also tried to compare it with various divergent series, but nothing worked...

Does anybody have any idea for a simple solution?

Answer 1

Note $$\frac{(n+1)!e^{n+1}/(n+1)^{n+1}}{n!e^n/n^n}=\frac{(n+1)e}{(n+1)^{n+1}/n^n}=\frac{e}{(n+1)^n/n}=\frac{e}{(1+1/n)^n}>1$$ since $e>(1+1/n)^n$ for all $n$. It follows that $$\frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}>\frac{n!e^n}{n^n},$$ so the terms of this series are increasing, hence do not tend to zero, thus the series is divergent.

Now, perhaps your teacher meant $\frac{n^n}{n!e^n}$, as then the series would still be divergent, but the terms do go to zero. However, there is a trick to make it work -- multiply the terms by $n$. By repeating a calculation like above, from $e<(1+1/n)^{n+1}$ it follows that the seqence $n\cdot\frac{n^n}{n!e^n}$ is increasing, so all its terms are at least as large as the first one, equal to $1/e$. Hence $\frac{n^n}{n!e^n}\geq\frac{1}{en}$, and by divergence of harmonic series this series diverges too.

Answer 2

It's may be interesting that $$\sum_{n=1}^{\infty} \frac{n!e^n}{n^{n+a}}$$ is converged when $a> \frac{3}{2}$, which can be obtained by Raabe's test

Answer 3

Using Rabee test (if $ \lim_{n\to\infty} n(\frac{a_{n+1}}{a_{n}}-1) <1 $ diverges)

so $\lim_{n\to\infty}n(\frac{a_{n+1}}{a_{n}}-1) = \lim_{n\to\infty}n(\frac{\frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}}{\frac{n!e^{n}}{n^{n}}}-1)= \lim_{n\to\infty}n(\frac{(n+1)!e^{n+1}n^{n}}{(n+1)^{n+1}n!e^{n}}-1)= \lim_{n\to\infty}n(\frac{(n+1)n^{n}e}{(n+1)^{n+1}}-1)= \lim_{n\to\infty}n(\frac{n^{n}e}{(n+1)^{n}}-1)=\lim_{n\to\infty}n(e{(\frac{n+1}{n})}^{-n}-1)=\frac{1}{2}$

so the serie converges (you can evaluate the last limit using laurent series)