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: 22



Work Step by Step

Given $$c_{n}=\frac{\ln \left(n^{2}+1\right)}{\ln \left(n^{3}+1\right)}$$ Then \begin{align*} \lim_{n\to \infty } c_n &= \lim_{n\to \infty } \frac{\ln \left(n^{2}+1\right)}{\ln \left(n^{3}+1\right)}\ \ \\ & \text{Using L'Hopital's rule } \\ &=\lim _{n\to \infty \:}\left(\frac{\frac{2n}{n^2+1}}{\frac{3n^2}{n^3+1}}\right)\\ &=\lim _{n\to \infty \:}\left(\frac{2\left(n^3+1\right)}{3n\left(n^2+1\right)}\right)\\ &=\lim _{n\to \infty \:}\left(\frac{2n^2}{3n^2+1}\right) \\ &= \lim _{n\to \infty \:}\left(\frac{4n}{6n}\right)\\ &=\frac{2}{3} \end{align*}
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