Answer
The series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{{n^2}}}{{{{\left( {{n^3} + 1} \right)}^{1.01}}}}$ converges.
Work Step by Step
To apply the Integral Test, we find the limits of integration by plotting the graph of the function $f\left( x \right) = \dfrac{{{x^2}}}{{{{\left( {{x^3} + 1} \right)}^{1.01}}}}$. From the figure attached, we see that $f$ is positive, decreasing, and continuous for $x \ge 2$. So, we evaluate the definite integral:
$\mathop \smallint \limits_2^\infty \dfrac{{{x^2}}}{{{{\left( {{x^3} + 1} \right)}^{1.01}}}}{\rm{d}}x$
Let $t = {x^3} + 1$. So, $dt = 3{x^2}dx$.
$\mathop \smallint \limits_2^\infty \dfrac{{{x^2}}}{{{x^3} + 1}}{\rm{d}}x = \dfrac{1}{3}\mathop \smallint \limits_9^\infty \dfrac{1}{{{t^{1.01}}}}{\rm{d}}t = \dfrac{1}{3}\left[ { - \dfrac{1}{{0.01}}{t^{ - 0.01}}} \right]_9^\infty = \dfrac{1}{3}\left( {\dfrac{1}{{0.01}}\cdot{9^{ - 0.01}}} \right) \approx 32.61$
Since $\mathop \smallint \limits_2^\infty \dfrac{{{x^2}}}{{{{\left( {{x^3} + 1} \right)}^{1.01}}}}{\rm{d}}x$ converges, by the Integral Test, the series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{{n^2}}}{{{{\left( {{n^3} + 1} \right)}^{1.01}}}}$ also converges.
