<s>[INST] Divergence of a Series $\sum_{n=1}^\infty (\frac{1}{n!})(\frac{n}{e})^n$ I'm trying to show that $$\sum_{n=1}^\infty \left(\frac{1}{n!}\right)\left(\frac{n}{e}\right)^n$$ diverges by the Corollary of Raabe's Test.

Corollary of Raabe's Test: Let $X=\{x_n\}$ be a sequence of nonzero real numbers and let $a=\lim_{n \to \infty} (n(1-\frac{|x_n+1|}{x_n}))$, whenever this limit exists. Then the series is absolutely convergent when $a>1$ and is divergent when $a<1$.

What I find:
$$a= \lim_{n \to \infty} (n(1-\frac{1}{e}(1+\frac{1}{n})^n))=\infty \cdot 0$$ (which is indeterminate)
$$\lim_{n \to \infty} (n(1-\frac{1}{e}(1+\frac{1}{n})^n))=\lim_{n \to \infty} \frac{1-\frac{1}{e}(1+\frac{1}{n})^n)}{\frac{1}{n}}$$
Applying L'hospital's rule, I get: $$\lim_{n \to \infty} \frac{-\frac{n}{e}(1+\frac{1}{n})^{n-1}(-\frac{1}{n^2})}{-\frac{1}{n^2}}=\lim_{n \to \infty} (-\frac{n}{e}(1+\frac{1}{n})^{n-1})$$
Help!!! I do not know what to do next.
 [/INST] The limit is $a=1/2$.
Start by the approximation
$$\ln \left( \frac 1e (1+x)^{1/x} \right) = -1 + \frac 1x \ln (1+x) = -1 + 1-\frac 12 x  +o(x) = -\frac 12 x  +o(x) $$
Plugging $x= \frac 1n$ you get
$$1- \frac 1e \left( 1+ \frac 1n \right)^n = 1- e^{\ln \left( \frac 1e (1+\frac 1n)^{n} \right)} = 1- e^{-\frac{1}{2n} + o( 1/n)}$$
Now, again, using the limit
$$\lim_{x \to 0} \frac{1-e^x}{x} = -1$$
we have
$$a= \lim_{n \to \infty} \frac{1- e^{-\frac{1}{2n} + o( 1/n)}}{1/n} =
\lim_{n \to \infty} \frac{1- e^{-\frac{1}{2n} + o( 1/n)}}{-\frac{1}{2n} + o( 1/n)} \cdot \frac{-\frac{1}{2n} + o( 1/n)}{1/n}
= -1 \cdot \left( -\frac{1}{2} \right) = \frac{1}{2}$$
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