Integrate using our knowledge of the power rule:
$$ \int\limits_1^\infty \frac{1}{x^{3p+1}} dx = \lim\limits_{b\to\infty} -\frac{1}{3p}\cdot\frac{1}{x^{3p}}\Big]_1^b $$
We can ignore the the coefficient since \(p\) is a constant.
$$ \lim_{b\to\infty} \frac{1}{b^{3p}}-\frac{1}{1^{3p}} $$
We can ignore the second term since it will evaluate to be a constant as well.
$$ \lim_{b\to\infty} \frac{1}{b^{3p}} $$
If the exponent is negative, the integral will diverge regardless of the value of the exponent.
$$ \lim_{b\to\infty} \frac{1}{b^{-a}} = \lim_{b\to\infty} b^a = \infty $$
If the exponent is positive, the denominator will approach infinity and the limit will evaluate to zero and therefore converge.
$$ \lim_{b\to\infty} \frac{1}{b^{a}} = 0 $$
For the exponent to be positive,
$$ 3p\gt 0 $$
$$ \boxed{p \gt 0} $$