Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.2 Summing and Infinite Series - Exercises - Page 547: 24



Work Step by Step

Given $$\frac{4^{3}}{5^{3}}+\frac{4^{4}}{5^{4}}+\frac{4^{5}}{5^{5}}+\cdots$$ Since the series is a geometric series with $|r|=\left|\frac{4}{5}\right|<1$ , then the series converges and has the sum \begin{align*} S&= \frac{a_1}{1-r}\\ &=\frac{\frac{4^{3}}{5^{3}}}{1-\frac{4}{5}}\\ &= \frac{4^{3}}{5^{2}}=\frac{64}{25} \end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.