College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.1 - Sequences and Summation Notation - 8.1 Exercises: 71

Answer

$\sum_{k=1}^{999}\frac{1}{k(k+1)}$

Work Step by Step

We write the sum in sigma notation: (We notice that the denominator takes the form of $n(n+1)$ with $n$ being consecutive integers.) $ \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots+\frac{1}{999\cdot 1000}=\sum_{k=1}^{999}\frac{1}{k(k+1)}$
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