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: 25

Answer

$$ 0.a_1a_2.......a_n+10^{-n} \times 1$$

Work Step by Step

We have: $0.a_1a_2.......a_n9999......=0.a_1a_2.......a_n+10^{-n} \times 0.9999......$ This shows a geometric series with common ratio $r=0.1$ and with initial term $a=0.9$. We see that $|r|<1$, so the series converges. Therefore, the sum is: $$0.9999=\dfrac{a}{1-r}\\=\dfrac{0.9}{1-0.1}\\=1$$ Therefore, the sum is: $$0.a_1a_2.......a_n9999......= 0.a_1a_2.......a_n+10^{-n} \times 1$$
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