I think the given series is
[tex]\displaystyle\sum_{n=2}^\infty \frac{n^2}{n^3+1}[/tex]
You can use the integral test because the summand is clearly positive and decreasing. Then
[tex]\displaystyle\sum_{n=2}^\infty\frac{n^2}{n^3+1} > \int_2^\infty\frac{x^2}{x^3+1}\,\mathrm dx[/tex]
Substitute u = x ³ + 1 and du = 3x ² dx, so the integral becomes
[tex]\displaystyle \int_2^\infty\frac{x^2}{x^3+1}\,\mathrm dx = \frac13\int_9^\infty\frac{\mathrm du}u = \frac13\ln(u)\bigg|_{u=9}^{u\to\infty}[/tex]
As u approaches infinity, we have ln(u) also approaching infinity (whereas 1/3 ln(9) is finite), so the integral and hence the sum diverges.