## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.5 The Ratio and Root Tests and Strategies for Choosing Tests - Exercises - Page 568: 57

Converges

#### Work Step by Step

Given$$\sum_{n=1}^{\infty} 4^{-2 n+1}=\sum_{n=1}^{\infty} 4\left(4^{-2}\right)^{n}$$ Then $$\sum_{n=1}^{\infty} 4^{-2 n+1}=\sum_{n=1}^{\infty} 4 \frac{1}{16^{n}}$$ which is a geometric series with $c=4$ and $r=\frac{1}{16}<1$. Thus, the given series converges.

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