Answer
$\sum ^{9}_{i=1}\dfrac {1}{3i}$
Work Step by Step
$\dfrac {1}{3\left( 1\right) }+\dfrac {1}{3\left( 2\right) }+\dfrac {1}{3\left( 3\right) }...=\dfrac {1}{3\times 1}+\dfrac {1}{3\times 2}+\dfrac {1}{3\times 3}+\ldots +\dfrac {1}{3\times 9}=\sum ^{9}_{i=1}\dfrac {1}{3i}$