Answer
The radius of convergence is $R=e$.
Work Step by Step
Ratio Test; Let us consider an infinite series $\Sigma a_k$ with positive terms and suppose that $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|$
1) When $l \lt 1$. then series $\Sigma a_k$ converges absolutely.
2) When $l \gt 1$. then series $\Sigma a_k$ diverges.
3) When $l = 1$. then series $\Sigma a_k$ may converge or diverge, that is, the test is inconclusive.
Here, we have $a_k=\dfrac{k ! \ x^k}{k^k}$
Now, $l=\lim\limits_{n \to \infty}|\dfrac{a_{k+1}}{a_k}|=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(k+1) ! \ x^{k+1}}{(k+1)^k}}{\dfrac{k ! \ x^k}{k^k}}|=|x| \times \lim\limits_{n \to \infty} |(\dfrac{k}{k+1})^k|=|x| \times \lim\limits_{n \to \infty} |(1-\dfrac{1}{k+1})^{k+1 \cdot \dfrac{k}{k+1}}|=|x| \times \dfrac{1}{e}$
This means that the series converges for $r=\dfrac{1}{e}|x| \lt 1$ . Then $-1 \lt \dfrac{1}{e} \cdot x \lt 1 \implies -e \lt x \lt e$
So, the radius of convergence for the power series centered at $0$ is $R=e$
