Calculus (3rd Edition)

$\frac{4}{e^2-2e}.$
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 $$\frac{c}{1-r}=\frac{4/e^2}{1-(2/e)}=\frac{4}{e^2-2e}.$$