Answer
$$2-1/e\lt x\lt 2+1/e$$
Work Step by Step
Given $$\sum_{n=12}^{\infty} e^{n}(x-2)^{n}$$ Since $a_n = e^{n}(x-2)^{n}$ and $a_{n+1} =e^{n+1}(x-2)^{n+1}$, then \begin{aligned} \rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|\frac{e^{n+1}(x-2)^{n+1}}{e^{n}(x-2)^{n}}\right|\\ &=\lim _{n \rightarrow \infty}|e(x-2)|\\ &=e|x-2| \end{aligned} Then the series converges for $$e|x-2|\lt 1 \ \to \ \ 2-1/e\lt x\lt 2+1/e$$
Now, we check the end points.
For $x= 2-1/e $ $$\sum_{n=12}^{\infty} e^{n}(x-2)^{n}=\sum_{n=1}^{\infty} (-1)^n $$
which diverges.
For $x= 2+1/e$ $$\sum_{n=12}^{\infty} e^{n}(x-2)^{n}=\sum_{n=1}^{\infty}(1)^n $$
which is divergent.
Hence, the interval of convergence is $$2-1/e\lt x\lt 2+1/e$$
