Answer
$\frac{4}{e^2-2e}.$
Work Step by Step
The series $\sum_{n=2}^{\infty}\left(\frac{2}{e}\right)^{n}$ can be rewritten as follows
$$
\sum_{n=2}^{\infty}\left(\frac{2}{e}\right)^{n}= \left(\frac{2}{e}\right)^{2}+\left(\frac{2}{e}\right)^{3}+\cdots \\
=\sum_{n=0}^{\infty}\frac{4}{e^2}\left(\frac{2}{e}\right)^{n}.
$$
Now, we have a geometric series which has the sum
$$\frac{c}{1-r}=\frac{4/e^2}{1-(2/e)}=\frac{4}{e^2-2e}.$$
