Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 12 - Sequences; Induction; the Binomial Theorem - Chapter Review - Review Exercises - Page 838: 6


$\sum_{k=1}^{13} (-1)^{k+1} (\frac{1}{k})$

Work Step by Step

We can see that this is the sum of the $13$ fractions in which the numerator is $1$, and in which the first denominator is $1$ and increases by $1$ in each term. Also, the signs are alternating, starting with $+$. Hence, the summation formula is: $\sum_{k=1}^{13} (-1)^{k+1} (\frac{1}{k})$
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