## College Algebra 7th Edition

$\frac{11}{6}$
We evaluate the sum: $\sum_{k=1}^{3} \frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{6}{6}+\frac{3}{6}+\frac{2}{6}=\frac{11}{6}$