University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises - Page 299: 13



Work Step by Step

$\frac{1}{2}=\frac{1}{2^1}, \frac{1}{4}=\frac{1}{2^2},\frac{1}{8}=\frac{1}{2^3},\frac{1}{16}=\frac{1}{2^4}$, therefore the kth element is $\frac{1}{2^k}$, therefore the formula is: $\sum\limits_{k=1}^4\frac{1}{2^k}$
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