Answer
The p-series converges
Work Step by Step
Use the integral test to determine convergence or divergence
$\Sigma^{\infty} _{n=1} \frac{1}{n^3}$
It is positive, and continuous on the interval $n\geq 1$
and it is decreasing on the interval because
$y' =-\frac{3}{n^4}$, which is negative for all numbers
Now the integral test can be used
$\int _1^{\infty} \frac{1}{n^3}dn$
$\lim\limits_{b \to \infty} [-\frac{1}{2n^2}]^b_1$
$\lim\limits_{b \to \infty} (-\frac{1}{2b^2} + \frac{1}{2})$
$\frac{1}{2}$, the p-series converges
