Answer is: option3
\( e \)Solution:
To determine the radius of convergence, evaluate \[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| \]
Let \[ a_n=\frac{n!x^n}{n^n} \]
Then \[ \left|\frac{a_{n+1}}{a_n}\right| = \left| \frac{(n+1)!x^{n+1}}{(n+1)^{n+1}} \cdot \frac{n^n}{n!x^n} \right| \]
\[ = \left| x\cdot\frac{n^n}{(n+1)^n} \right| = \left| \frac{x}{\left(1+\frac{1}{n}\right)^n} \right| \]
Using \[ \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e \] we get \[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x}{e}\right| \]
The series converges if \[ \left|\frac{x}{e}\right|<1 \]
Hence
\[
|x|
Therefore the radius of convergence is
\[
R=e
\]
