Calculus 8th Edition

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

Appendix E - Sigma Notation - E Exercises: 44

Answer

$\frac{5}{4}$

Work Step by Step

Find the limit $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}[(\frac{i}{n})^{3}+1]$ $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}[(\frac{i}{n})^{3}+1]=\lim\limits_{n \to \infty}(\frac{1}{n^{4}}\sum\limits_{i =1}^{n}{i}^{3}+\frac{1}{n}\sum\limits_{i =1}^{n}1)$ Since, $ \sum \limits_{i =1}^{n}i^{3}=[\frac{n(n+1)}{2}]^{2}$ Thus, $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}[(\frac{i}{n})^{3}+1]=\lim\limits_{n \to \infty}(\frac{1}{n^{4}}[\frac{n(n+1)}{2}]^{2}+\frac{1}{n}(n)$ $=\lim\limits_{n \to \infty}[\frac{1}{4}+\frac{1}{2n}+\frac{1}{4n^{2}}+1]$ $=\frac{1}{4}+0+0+1$ Hence, $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}[(\frac{i}{n})^{3}+1]=\frac{5}{4}$
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