## Multivariable Calculus, 7th Edition

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