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: 17



Work Step by Step

Since$$b_{n}=\sqrt{n^{2}+n}-\sqrt{n^{2}+1}$$ Then \begin{align*} \lim _{n \rightarrow \infty} b_{n}&=\lim _{n \rightarrow \infty} (\sqrt{n^{2}+n}-\sqrt{n^{2}+1})\frac{(\sqrt{n^{2}+n}+\sqrt{n^{2}+1})}{\sqrt{n^{2}+n}+\sqrt{n^{2}+1}}\\ &= \lim _{n \rightarrow \infty} \frac{\left(n^{2}+n\right)-\left(n^{2}+1\right)}{\sqrt{n^{2}+n}+\sqrt{n^{2}+1}}\\ &=\lim _{n \rightarrow \infty} \frac{n-1}{\sqrt{n^{2}+n}+\sqrt{n^{2}+1}} \\ &=\lim _{n \rightarrow \infty} \frac{\frac{n}{n}-\frac{1}{n}}{\sqrt{\frac{n^{2}}{n^{2}}+\frac{n}{n^{2}}}+\sqrt{\frac{n^{2}}{n^{2}}+\frac{1}{n^{2}}}}\\ &=\lim _{n \rightarrow \infty} \frac{1-\frac{1}{n}}{\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{1}{n^{2}}}}\\ &=\frac{1}{2} \end{align*}
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