Answer
$$ -\infty\lt x\lt \infty $$
Work Step by Step
Given $$\sum_{n=0}^{\infty} \frac{(x-4)^{n}}{n !}$$ Since $a_n = \frac{(x-4)^{n}}{n !}$ and $a_{n+1} = \frac{(x-4)^{n+1}}{(n+1) !}$, then \begin{aligned} \rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ & =\lim _{n \rightarrow \infty}\left|\frac{\frac{(x-4) (x-4)^{n}}{(n+1) n !}}{\frac{(x-4)^{n}}{n !}}\right|\\ & =\lim _{n \rightarrow \infty}\left|(x-4) \frac{1}{n+1}\right|\\ & =|(x-4)| \lim _{n \rightarrow \infty} \frac{1}{n+1}\\ &=0 \end{aligned}
Then the interval of convergence is
$$ -\infty\lt x\lt \infty $$
