Determine the radius of convergence 
of the power series:
![\[ \sum_{n=1}^{\infty} \frac{n! z^n}{n^n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4834583291ae1b66cc07349b00d59e4d_l3.png "Rendered by QuickLaTeX.com")
Test for convergence at the boundary points if is finite.
Let 
. Then,
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(n+1)! z^{n+1}}{(n+1)^{n+1}} \right) \left( \frac{n^n}{n! z^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{z n^n (n+1)}{(n+1)^{n+1}} \\[9pt] &= \lim_{n \to \infty} \left( \frac{zn^{n+1}}{(n+1)^{n+1}} + \frac{zn^n}{(n+1)^{n+1}} \right) \\[9pt] &= \lim_{n \to \infty} \left( \frac{n}{n+1} \cdot \left( \frac{n}{n+1} \right)^n z \right) \\[9pt] &= 1 \cdot \frac{1}{e} \cdot z \\[9pt] &= \frac{z}{e}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f0d10fe6040c0960844072ed38ead050_l3.png "Rendered by QuickLaTeX.com")
Thus, the series converges if 
which implies 
or the radius of convergence is 
. It diverges on the boundary points since the terms do not go to 0.