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 592: 54



Work Step by Step

Given $$\sum_{n=2}^{\infty}(1-\sqrt{1-\frac{1}{n^{2}}}) $$ Since \begin{aligned} a_{n}&=1-\sqrt{1-\frac{1}{n^{2}}}\\ & =1-\sqrt{\frac{n^{2}-1}{n^{2}}}\\ & =1-\frac{\sqrt{n^{2}-1}}{n}\\ & =\frac{n-\sqrt{n^{2}-1}}{n}\\ & =\frac{n-\sqrt{n^{2}-1}}{n} \times \frac{n+\sqrt{n^{2}-1}}{n+\sqrt{n^{2}-1}}\\ &= \frac{1}{n^{2}+n \sqrt{n^{2}-1}} \end{aligned} Compare with the convergent series $\sum \frac{1}{n^2}$: \begin{align*} \lim_{n\to \infty } \frac{a_n}{b_n} &= \lim_{n\to \infty } \frac{n^2}{n^{2}+n \sqrt{n^{2}-1}}\\ &=1 \end{align*} Then $\sum_{n=2}^{\infty}(1-\sqrt{1-\frac{1}{n^{2}}})$ converges.
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