## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - Chapter Review Exercises - Page 591: 35

#### Answer

$$a_{n}=-1, \ \ \ \ b_{n}=\left(\frac{1}{2}\right)^{n}+1$$

#### Work Step by Step

Consider $$a_{n}=-1, \ \ \ \ b_{n}=\left(\frac{1}{2}\right)^{n}+1$$ Since \begin{align*} \lim_{n\to \infty } a_n &=-1\neq 0\\ \lim_{n\to \infty } b_n &= 1\neq 0 \end{align*} Then $\sum a_n,\ \ \sum b_n$ are divergent series, but $$\sum_{n=1}^{\infty} (a_n+b_n)=\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n}$$ which is a geometric series with $r= 1/2$ and has the sum $$\frac{a}{1-r}= \frac{\frac{1}{2}}{1-\frac{1}{2}}=1$$

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