Answer
$\frac{7}{6}$
Work Step by Step
$\sum\limits_{k=1}^3\frac{k-1}{k}=\frac{1-1}{1}+\frac{2-1}{2}+\frac{3-1}{3}=\frac{0}{1}+\frac{1}{2}+\frac{2}{3}=0+\frac{3}{6}+\frac{4}{6}=\frac{7}{6}$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
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: 2
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