## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises: 15

#### Answer

$\sum\limits_{k=1}^5(-1)^{k+1}\cdot\frac{1}{k}$

#### Work Step by Step

As we can see, the fraction's denominator is always growing by one, and the fraction's sign is always changing, therefore the formula is: $\sum\limits_{k=1}^5(-1)^{k+1}\cdot\frac{1}{k}$

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