## College Algebra (11th Edition)

$\sum_{i=1}^9(\frac{1}{3i})$ or $\frac{1}{3}\sum_{i=1}^{9}(\frac{1}{i})$
We can see that the number in the denominator, in brackets is changing from 1 to 9, with steps of 1. Therefore this will be our variable, and the $\frac{1}{3}$ is constant in all the fractions in the series. We can write that the sum is: $\sum_{i=1}^9(\frac{1}{3i})=\frac{1}{3}\sum_{i=1}^{9}\frac{1}{i}$