How do I prove whether the series converges when $a_n = \frac {n! e^n}{n^n}$?

Question

The series $\sum a_n$ is the series I get when analyzing the endpoint of a power series, so the ratio test is out of the question. How else can I find whether this particular series converges?

Answer 1

What's below is overkill, as pointed out in the comments. Indeed, once it's established that $$ \frac{a_{n+1}}{a_n} = \frac{e}{(1+\frac{1}{n})^{n}} > 1 $$ then it's clear that the sequence $(a_n)_n$ is inctreasing. Therefore, $a_n \not\to 0$, and the series $\sum_n a_n$ trivially diverges by the limit test.

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Since Stirling's approximation is not allowed*, let's try extensions of the ratio test.

Since $$ \frac{a_{n+1}}{a_n} = \frac{(n+1)e\cdot n^n}{(n+1)^{n+1}} = \frac{e\cdot n^n}{(n+1)^{n}} = \frac{e}{(1+\frac{1}{n})^{n}} $$ we have $$\begin{align*} n\left( \frac{a_n}{a_{n+1}} -1 \right) &= e^{-1}n\left( \left(1+\frac{1}{n}\right)^{n} - e\right) = e^{-1}n\left( e^{n \ln\left(1+\frac{1}{n}\right)} - e\right)\\ &= e^{-1}n\left( e^{n(\frac{1}{n} - \frac{1}{2n^2}+o\left(\frac{1}{n^2}\right))} - e\right)\\ &= e^{-1}n\left( e^{1-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)} - e\right) = n\left( e^{-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)} - 1\right)\\ &= n\left( -\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\right)\\ &\xrightarrow[n\to\infty]{} \boxed{-\frac12} < 1 \end{align*}$$ so the series diverges by Raabe's test.

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${}^\ast$ Of course, if you want to use it: $n!\displaystyle\operatorname*{\sim}_{n\to\infty} {\sqrt{2\pi n}}\frac{n^n}{e^n}$, so $$ a_n \operatorname*{\sim}_{n\to\infty} {\sqrt{2\pi n}} $$ and the series $\sum_n a_n$ (of positive terms) diverges by the limit test.