#### Answer

Divergent

#### Work Step by Step

A sequence is said to be converged if and only if $\lim\limits_{n \to \infty}a_{n}$ is a finite constant.
$\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}\frac{n^{3}}{1+n^{2}}$
Divide numerator and denominator by $n^{3}$.
$\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}\frac{\frac{n^{3}}{n^{3}}}{\frac{1+n^{2}}{n^{3}}}$
$=\lim\limits_{n \to \infty}\frac{1}{\frac{1}{n^{3}}+\frac{1}{n}}$
$=\frac{1}{0+0}$
$=\infty$
We can also use L-Hospital's rule because the limit is of the form $\frac{\infty}{\infty}$
After applying L-Hospital's rule we get
$\lim\limits_{n \to\infty}\frac{3n^{2}}{2n}=\lim\limits_{n \to\infty}\frac{3n}{2}=\infty$
Hence, the given sequence is divergent.