Answer
Convergent
Work Step by Step
To use the Direct Comparison Test, we need to known series $\Sigma_{n=1}^{\infty}b_{n}$ for the purpose of comparison.
We will choose $b_{n}=\frac{2}{n^{2}}$
The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$.
Now, $a_{n}=\frac{n!}{n^{n}}$ and $b_{n}=\frac{2}{n^{2}}$; which is a p-series with $p =2\gt 1$
Here, $a_{n} \lt b_{n}$ for $n\geq 4$ , because $\frac{1}{n}\times \frac{2}{n} \times [\frac{3}{n} \times \frac{4}{n}...\frac{n}{n}] \lt \frac{1}{n} \times \frac{2}{n}$
Therefore, $\Sigma a_{n}$ is convergent by the comparison test.
Hence, the $\Sigma_{n=1}^{\infty}\frac{n!}{n^{n}}$ series is convergent.
