#### Answer

The given series is divergent.

#### Work Step by Step

$\frac{1}{3} +\frac{1}{6} +\frac{1}{9} +\frac{1}{12} + \frac{1}{15} +...$
$=\Sigma^{\infty}_{n=1} \frac{1}{3n}$
$= \frac{1}{3} \Sigma^{\infty}_{n=1} \frac{1}{n}$
We know that the harmonic series $\Sigma^{\infty}_{n=1} \frac{1}{n}$ is divergent.
Hence the series is divergent.