#### Answer

Divergent

#### Work Step by Step

$\Sigma \frac{2}{5n-1}$
$\int_{1}^{\infty}\frac{2}{5x-1}dx=\lim\limits_{t \to \infty}\int_{1}^{t}\frac{2}{5x-1}dx=\lim\limits_{t \to \infty}[\frac{2}{5}ln(5x-1)]_{1}^{t}=\infty-\frac{2}{5}ln(4)=\infty$
Divergent