Answer
Divergent
Work Step by Step
Find the limit of this sequence when $n\to\infty$ as:
$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{n^3}{2n+6}=\lim_{n\to\infty}\frac{n^3+3n^2-3n^2-9n+9n}{2(n+3)}=\lim_{n\to\infty}\frac{n^2(n+3)-3n(n+3)+9n}{2(n+3)}=\lim_{n\to\infty}(\frac{n^2}{2}-\frac{3n}{2}+\frac{9n}{2(n+3)})$
Clearly, this sequence is divergent when $n\to \infty$
