## Calculus (3rd Edition)

$$\frac{e}{e^2-1}$$
Given $$\sum_{n=2}^{n=\infty} e^{3-2 n}=\sum_{n=2}^{n=\infty} e^{3}\left(\frac{1}{e^{2}}\right)^{n}$$ Since the series is a geometric series with $|r|=\left|\frac{1}{e^2}\right|<1$ , then the series converges and has the sum \begin{align*} S&= \frac{a_1}{1-r}\\ &=\frac{\frac{1}{e}}{1-\frac{1}{e^2}}\\ &=\frac{e}{e^2-1} \end{align*}