Answer
The series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{{n^2}}}{{{n^3} + 1}}$ does not converge.
Work Step by Step
To apply the Integral Test, we find the limits of integration by plotting the graph of the function $f\left( x \right) = \dfrac{{{x^2}}}{{{x^3} + 1}}$. From the figure attached, we see that $f$ is positive, decreasing, and continuous for $x \ge 2$. So, we evaluate the definite integral:
$\mathop \smallint \limits_2^\infty \dfrac{{{x^2}}}{{{x^3} + 1}}{\rm{d}}x$
Let $t = {x^3}$. So, $dt = 3{x^2}dx$.
$\mathop \smallint \limits_2^\infty \dfrac{{{x^2}}}{{{x^3} + 1}}{\rm{d}}x = \dfrac{1}{3}\mathop \smallint \limits_8^\infty \dfrac{1}{{t + 1}}{\rm{d}}t = \dfrac{1}{3}\left[ {\ln \left( {t + 1} \right)} \right]_8^\infty = \infty $
Since $\mathop \smallint \limits_2^\infty \dfrac{{{x^2}}}{{{x^3} + 1}}{\rm{d}}x$ does not converge, by the Integral Test, the series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{{n^2}}}{{{n^3} + 1}}$ does not converge.
