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