Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

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



Work Step by Step

We have $$ \lim _{n \rightarrow \infty} a_n=\lim _{n \rightarrow \infty}2^{1/n^2}= 2^0=1. $$ Hence, $a_n$ converges to $1$.
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