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The idea is to show that $\sum\limits_{n=0}^{\infty} \dfrac{n^n}{n!e^n}$ diverges, but that $\lim_{n \to\infty}\dfrac{n^n}{n!e^n} = 0$ (which is the reason why the series is challenging).

I was slogging away at this and eventually arrived at some pretty short elementary proofs, but was also surprised to have seen no discussion of this on SE, so I'm attaching my answer as well so that people can see if I've overlooked something crucial.

Another, less precise question: would this be considered a rather 'tight' or 'unobvious' divergence? What would be a very difficult series for which all the usual tests (ratio, root, eyeball comparison, $2^na_{2^n}$, Raabe's, Taylor expansion etc.) fail?

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  • $\begingroup$ Shouldn't your sum start at $1$ since $0^0$ is undefined $\endgroup$
    – JasonM
    Jun 8 '16 at 8:36
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Stirling's approximation for $n!$ says $n!$ can be estimated by $\sqrt{2\pi n}(\frac{n}{e})^n$ for large values. The terms of your series are therefore asymptotic to terms of the form $\frac{1}{\sqrt{2\pi n}}$. Since the sum $\sum_{n=1}^{\infty} \frac{1}{\sqrt{2\pi n}}$ is a divergent $p$-series, we know your sum is divergent as well.

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  • $\begingroup$ I meant fast but elementary, but I suppose Stirling's will do. $\endgroup$
    – Yon Teh
    Jun 8 '16 at 8:15
  • $\begingroup$ @ChrisSanders Well you could use a weaker bound if you like. For example the wiki for factorial under rate of growth gives a couple weak bounds. $\endgroup$
    – JasonM
    Jun 8 '16 at 8:35
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We may use the Lambert $W$ function. The radius of convergence of $$W(x) = \sum_{n\geq 1}\frac{(-n)^{n-1}}{n!}\,x^n $$ is $\frac{1}{e}$ by the ratio test, and

$$ \sum_{n\geq 1}\frac{n^n}{n!}\,x^n = x\cdot\frac{d}{dx}\left(-W(-x)\right)=\color{red}{-1+\frac{1}{1+W(-x)}} $$

cannot converge at $x=\frac{1}{e}$, since $W\left(-\frac{1}{e}\right)=-1$.


That also leads to a quite unusual representation for $1$: $$ 1 = \sum_{n\geq 1}\frac{n^n}{n!(4e)^{n/2}}.$$

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    $\begingroup$ (+1). Another closely related series is $$\sum_{n=1}^{\infty} \left( \frac{n^n}{n! e^n}-\frac1{\sqrt{2\pi n}}\right)=-\frac23-\frac1{\sqrt{2\pi}}\zeta\left(\frac12\right)$$ which is known as Knuth's series. $\endgroup$
    – nospoon
    Jun 8 '16 at 10:39
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1) It suffices to observe that $\dfrac{n^n}{n!e^n}>\dfrac{1}{en}$.

Indeed, let $a_n=\dfrac{n^{n+1}}{n!e^n}$. Then $\dfrac{a_{n+1}}{a_n}=\dfrac{1}{e}(1+\dfrac{1}{n})^{n+1}$, which is known to be greater than 1. Hence $a_n>a_1=\dfrac{1}{e}$.

Note: for the original question, the ratio test is inconclusive, and the root test yields an ugly expression.

2) Now $e^n=\displaystyle \sum_{k=0}^\infty \frac{n^k}{k!}$, so evidently $e^n>\dfrac{n^n}{n!}$, but to go beyond this is a little subtler.

We must compare the $\dfrac{n^k}{k!}=a_k$. Since $\dfrac{a_{k+1}}{a_{k}}=\dfrac{n}{k+1}$,

$a_0<a_1<...<a_{n-1}=a_n$ and $ a_n>a_{n+1}>a_{n+2}>...$

We must try to show that for all $m∈\Bbb N$ and for any $\epsilon>0$ there exists $n$ such that $\dfrac{a_{n+m}}{a_n}>1-\epsilon$, so that $a_n+a_{n+1}+...+a_{n+m}>ma_n(1-\epsilon)$, whence $e^n$ is arbitrarily larger than $\dfrac{n^n}{n!}$.

It so happens that $\dfrac{a_{n+m}}{a_n}=(\dfrac{n}{n+1})(\dfrac{n}{n+2})...(\dfrac{n}{n+m})>(\dfrac{n}{n+m})^m$ and for large enough $n$, $\dfrac{n}{n+m}>(1-\epsilon)^{1/m}$ because m is fixed.

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