Perhaps it suffices to say that
[tex]\displaystyle\frac1{1^2}+\cdots+\frac1{n^2}=\sum_{k=1}^n\frac1{k^2}[/tex]
converges to [tex]\frac{\pi^2}6[/tex] as [tex]n\to\infty[/tex], while [tex]n^3\to\infty[/tex]. Then
[tex]\displaystyle\lim_{n\to\infty}a_n=\infty\cdot\frac{\pi^2}6=\infty[/tex]
so [tex]a_n[/tex] diverges.
