9. Find the positive constant $c_{0}$ such that the series
$$
\sum_{n=0}^{\infty} \frac{n!}{(c n)^{n}}
$$
converges for $c>c_{0}$ and diverges for $0<c<c_{0}$.
Solution:
```
1/e
```
The ratio test tells us that the series converges if
$$
\lim _{n \rightarrow \infty} \frac{(n+1)!/(c(n+1))^{n+1}}{n!/(c n)^{n}}=\frac{1}{c} \cdot \lim _{n \rightarrow \infty}\left(\frac{n}{n+1}\right)^{n}
$$
is less than one and diverges if it is greater than one. But
$$
\lim _{n \rightarrow \infty}\left(\frac{n}{n+1}\right)^{n}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{-n}=\frac{1}{e} .
$$

Then the limit above is just $1 / c e$, so the series converges for $c>1 / e$ and diverges for $0<c<1 / e$.