Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.4 - Comparison Tests - Exercises 10.4 - Page 592: 60



Work Step by Step

Given: $\lim\limits_{n \to \infty} n^2 a_n=0$ Let us consider that there exists a natural number $N$ such that $0 \lt n^2 a_n \leq 1$ for all $n \geq N$ This implies that when we compare it with: $\Sigma_{n=N}^\infty a_n \leq \Sigma_{n=N}^\infty \dfrac{1}{n^2}$, we find that the series $\Sigma_{n=N}^\infty a_n $ converges by the Limit comparison test. Therefore, the series $\Sigma _{n=1}^\infty a_n$ converges.
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