Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.4 Exercises - Page 751: 39

Answer

$\Sigma a^{2}_{n}$ also converges by the comparison test.

Work Step by Step

Since $\Sigma a_{n}$ converges, $\lim\limits_{n \to \infty}a_{n}=0$, so there exists N such that $|a_{n}-0|\lt 1$ for all $n\gt N \to 0\leq a_{n} \lt 1$ For all $n\gt N \to 0\leq a^{2}_{n} \lt a_{n}$. Since $\Sigma a_{n}$ converges, $\Sigma a^{2}_{n}$ also converges by the comparison test.
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