Proving the divergence of $\sum_{n=1}^\infty (-1)^n\frac{n^n}{n!}$

Question

I was asked to show whether

$$\sum_{n=1}^\infty (-1)^n\frac{n^n}{n!}$$

converges or diverges using Leibniz criterion if possible. I was wondering whether my proof of its divergence is correct.

Proof. Let $a_n= (-1)^n\frac{n^n}{n!}$ and $b_n = |a_n| =\frac{n^n}{n!}$

$\alpha$) The alternating series test can not be used because

$$b_{n+1} \leq b_n \iff (n+1)^{n+1}n! \leq n^n(n+1)!$$ $$\iff (n+1)^{n+1} \leq n^n(n+1)$$ $$\iff (n+1)^n \leq n^n$$

which can not be satisfied with $n \in \mathbb{N}$.

$\beta)$ Notice that $a_{2n} = \frac{(2n)^{2n}}{(2n)!}$ is a divergent subsequence of $a_n$, because

$$\lim_{n\to\infty} a_{2n} = \lim_{u\to\infty}a_u= \lim_{u\to\infty} \frac{u^u}{u!} \tag{$u=2n$} $$

$$= \lim_{u\to\infty} (\frac{u}{u} \frac{u}{u-1} \frac{u}{u-2} ... \frac{u}{2} \space u)$$

It is clear that the limit diverges because each factor $\frac{u}{u-a} \geq 1$ for $a \in [0, u-1]$.

$\gamma)$ Because a subsequence of $a_n$ diverges, then $a_n$ diverges; and because $a_n$ diverges, the series $\sum a_n$ diverges as well.

Is this proof correct? I am mostly worried about $\beta)$, when I state "it is clear that the limit diverges because...". One must be suspicious of statements of the kind "it is obvious $x$ occurs", but I could not find a more formal way to show the divergence of the subsequence.

NOTE: I believe my question differs from this one since I am not asking how to prove the divergence of the series. I am specifically asking whether my proof is correct and, if not, why.

Answer 1

You've already had the right idea for the proof, although your proof of "divergence" is not quite complete. However, remember that in order for a series to converge, its terms must converge to $0$, and $$b_n = |a_n| \ge 1 \quad\text{for all }n.$$

Answer 2

$$ \lim_{n -> \infty} \frac{n^{n}}{n!} = \lim_{n -> \infty} \frac{n^{n}}{\sqrt{2\pi n} n^{n}\frac{1}{e^{n}}} $$ by Stirling's approximation. You can see the limit diverges. Effectively you are summing up elements that get further and further away from $0$, let alone get to $0$ fast enough. So the sum diverges. The signs alternate but it only means that it is a different type of divergence (like the one for $e^{x} \sin x$ as $x$ goes to $\infty$), rather than one that allows you to deduce if the sum goes to $\infty$ or $-\infty$.