Answer
Divergent
Work Step by Step
$\frac{1}{3}+\frac{1}{7}+\frac{1}{11}+\frac{1}{15}+\frac{1}{19}+...=\Sigma \frac{1}{4n-1}$
$\int_{1}^{\infty}\frac{1}{4x-1}=\lim\limits_{t \to \infty}\int_{1}^{\infty} \frac{1}{4x-1}=\lim\limits_{t \to \infty}[\frac{1}{4}ln(4x-1)]^{t}_{1}=\infty-\frac{1}{4}ln(3)=\infty$
Divergent
