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: 10


$a_n$ converges to $-\frac{3}{2}$.

Work Step by Step

We have $$ \lim _{n \rightarrow \infty} a_n=\lim _{n \rightarrow \infty} \frac{3n^3-n}{1-2n^3} \\ =\lim _{n \rightarrow \infty} \frac{3-(1/n^2)}{(1/n^3)-2 } =\frac{3-0}{0-2}=-\frac{3}{2}. $$ Hence, $a_n$ converges to $-\frac{3}{2}$.
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