Answer
$\Sigma a_{n}$ is divergent.
Work Step by Step
Given that $a_{n}\gt 0$ , therefore, we can apply limit comparison test with $b_{n}=\frac{1}{n}$
Since, $\Sigma_{n=0}^{\infty}b_{n}=\Sigma_{n=0}^{\infty}\frac{1}{n}$ diverges
If $\lim\limits_{n\to \infty}\frac{a_{n}}{b_{n}}\ne 0$
Then according to limit comparison test $\Sigma_{n=0}^{\infty}a_{n}$ will also diverge.
$\lim\limits_{n\to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n\to \infty}\frac{a_{n}}{1/n}=\lim\limits_{n\to \infty}na_{n}$
$\ne 0$
Hence, $\Sigma a_{n}$ is divergent.
