Answer
$1$
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 $1$.
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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