Convergence of $\sum_{n=1}^{\infty}\frac{{e^n}{n!}}{n^n}$

Question

I tried to investigate the convergence of this series:
$$\sum_{n=1}^{\infty}\frac{{e^n}{n!}}{n^n}$$
I used ratio test and root test to test it at first, because they can solve another series that is similar to this:
$$\sum_{n=1}^{\infty}\frac{{2^n}{n!}}{n^n}$$

But I eventually found that it doesn't work, because they both result in inconclusive case ($C=1$).

Ratio test:
$$C=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{e(n+1)}{(n+1)(1+\frac{1}{n})^n}=1$$

Root test:
$$C=\lim_{n\to\infty}a_n^\frac{1}{n}=e\lim_{n\to\infty}(\frac{n!}{n^n})^\frac{1}{n}=e\lim_{n\to\infty}\exp{(\frac{1}{n}(\ln{\frac{1}{n}}+\ln{\frac{2}{n}}+\cdots+\ln{\frac{n}{n}}))}$$
$$=e\cdot\exp{(\int_0^1\ln{x}\,dx)}=1$$

Is there a easy (elementary?) way to test its convergence? Preferably without using stirling formula.

Answer 1

The ratio test in the form:

Let $(a_n)$ be a sequence of (strictly) positive terms. If $$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} < 1\,,$$ then the series $\sum a_n$ converges. If $$\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} > 1$$ or there is an $n_0$ such that $\frac{a_{n+1}}{a_n} \geqslant 1$ for $n \geqslant n_0$, then the series diverges (since the terms don't converge to $0$). Otherwise the test is inconclusive.

yields the divergence of the series, since the terms don't converge to $0$ (because $\bigl(1 + \frac{1}{n}\bigr)^n < e$ for all $n$).

The often-taught form of the ratio test that assumes the existence of $\lim \frac{a_{n+1}}{a_n}$ and only considers the value of the limit isn't strong enough to decide this case.