Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 9 - Infinite Series - 9.3 Infinite Series - Exercises Set 9.3 - Page 622: 24

Answer

$$\dfrac{44663}{99000}$$

Work Step by Step

We have: $0.451141414......=0.451+0.00014+......=0.451+\Sigma_{n=1}^{\infty} 0.00014 (0.01)^{n-1}$ This shows a geometric series with common ratio $r=0.01$ and with initial term $a=0.00014$. We see that $|r| \lt 1$, so the series converges. Therefore, the sum is: $$5.373737=0.451+\dfrac{a}{1-r}\\=0.451+\dfrac{0.00014}{1-0.01}\\=\dfrac{44663}{99}$$
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