Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix E - Sigma Notation - E Exercises - Page A38: 43



Work Step by Step

Find the limit $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}(\frac{i}{n})^{2}$ $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}(\frac{i}{n})^{2}=\lim\limits_{n \to \infty}\frac{1}{n^{3}}\sum\limits_{i =1}^{n}({i}^{2})$ Since, $ \sum \limits_{i =1}^{n}i^{2}=[\frac{n(n+1)(2n+1)}{6}]$ Thus, $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}(\frac{i}{n})^{2}=\lim\limits_{n \to \infty}[\frac{1}{n^{3}}(\frac{n(n+1)(2n+1)}{6})]$ $=\lim\limits_{n \to \infty}[\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^{2}}]$ $=\frac{1}{3}+0+0$ Hence, $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}(\frac{i}{n})^{2}=\frac{1}{3}$
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