Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - 11.3 Convergence of Series with Positive Terms - Exercises - Page 557: 54


The series $\Sigma_{n=1}^{\infty} \frac{1}{n^2+\sin n}$ converges.

Work Step by Step

Using the limit comparison test by $b_n=\frac{1}{n^2}$, now we have $$L=\lim_{n\to \infty} \frac{a_n}{b_n}= \lim_{n\to \infty} \frac{1/(n^2+\sin n)}{1/n^2}\\ =\lim_{n\to \infty} \frac{n^2}{n^2+\sin n}=\lim_{n\to \infty} \frac{1}{1+(\sin n/n^2)}=1$$ Since $\Sigma_{n=1}^{\infty}\frac{1}{n^2}$ is a convergent p-series and $L=1\gt 0$ then the series $\Sigma_{n=1}^{\infty} \frac{1}{n^2+\sin n}$ converges.
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