## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.2 Summing and Infinite Series - Exercises - Page 547: 30

#### Answer

$$\frac{e}{e^2-1}$$

#### Work Step by Step

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*}

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