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: 53



Work Step by Step

Since \begin{align*} \lim_{n\to \infty} a_n &=\lim_{n\to \infty } 1-\sqrt{1-\frac{1}{n}} \\ &=0 \end{align*} and \begin{align*} a_n &= 1-\sqrt{1-\frac{1}{n}} \\ &=1-\sqrt{\frac{n-1}{n}}\\ &=\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}}\\ &=\frac{n-(n-1)}{\sqrt{n}(\sqrt{n}+\sqrt{n-1})}\\ &=\frac{1}{n+\sqrt{n^{2}-n}} \\ & \geq \frac{1}{n+\sqrt{n^{2}}}\\ &\geq \frac{1}{2 n} \end{align*} Since $\sum \frac{1}{2 n}$ diverges, then by using the comparison test, $\sum a_n $ also diverges.
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