Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set - Page 1048: 40



Work Step by Step

Here, we have $\sum_{i=0}^4\dfrac{(-1)^{i+1}}{(i+1)!}=\dfrac{(-1)^{0+1}}{(0+1)!}+\dfrac{(-1)^{1+1}}{(1+1)!}+\dfrac{(-1)^{2+1}}{(2+1)!}+\dfrac{(-1)^{3+1}}{(3+1)!}+\dfrac{(-1)^{4+1}}{(4+1)!}$ $=-\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{24}-\dfrac{1}{120}$ $=\dfrac{-120+60-20+5-1}{120}$ $=\dfrac{-19}{30}$
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