## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 12 - Section 12.1 - Sequences and Summation Notation - 12.1 Exercises - Page 851: 70

#### Answer

sum = $\displaystyle \sum_{k=1}^{100}\frac{(-1)^{k+1}}{k\ln k}$

#### Work Step by Step

The kth term $a_{k}$ has $k\cdot\ln k$ in the denominator and $\pm 1$ in the numerator, (odd k's are minus, even k's are plus), so we write $(-1)^{k+1}$ for the numerator. (when k is even, the exponent is odd, and when k is odd, the exponent is even) $a_{k}=\displaystyle \frac{(-1)^{k+1}}{k\ln k}$ There are 100 terms (k=1,2,3,...,100). sum = $\displaystyle \sum_{k=1}^{100}\frac{(-1)^{k+1}}{k\ln k}$

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