# How to show that $\sum_{n=1}^\infty (-1)^n \frac{n^n}{n!}$ diverges?

I know that this series diverges.

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

From what I learned, as it's an alternating series, I have to prove that $$\frac{n^n}{n!}$$ descends and $$\lim_{n\to \infty} \frac{n^n}{n!} = 0$$.

However, I don't know what method I should use to break down $$\frac{n^n}{n!}$$.

(Also it's my first time editing formally like this, so if there's any mistake, any correction is appreciated!)

Since $$n^n>n!$$ as easily seen by induction you have that if you call with $$a_n$$ the general term of your series, it does not have limit $$0$$ as $$n \to \infty$$.Therefore the series is divergent.

HINT. Apply the Ratio Test carefully.

As a general note, some series which are alternating, are much easier shown to be convergent using the Root/Ratio Test (at least those which are absolutely convergent).

Stirling's formula for $$n!$$ is $$n! \approx \sqrt{2\pi} \, n^{n+1/2} e^{-n}$$

So the $$n^\textrm{th}$$ term of your series is approximately

$$a_n \approx (-1)^n \frac{e^n}{\sqrt{2\pi n} }.$$

The root test gives $$|a_n|^{1/n} \rightarrow e >1.$$

So the series diverges.

A necessary condition for any series $$\sum_{n=1}^\infty a_n$$ to converge is that $$\lim_{n\to\infty} |a_n| =0$$.

In your case is $$a_n = (-1)^n \frac{n^n}{n!}$$ and it is easy to verify that this sequence does not converge. For $$n >1$$ you have

$$|a_n| = \frac{\overbrace{n \cdots n}^{\lfloor \frac n2 \rfloor\; factors} \cdot n \cdots n}{1 \cdots \lfloor \frac n2 \rfloor \cdot (\lfloor \frac n2 \rfloor +1) \cdots n} > 2^{\lfloor \frac n2 \rfloor}\stackrel{n\to \infty}{\longrightarrow}+\infty$$ So, your series does not converge.