Answer
The series $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{\left( {n + 2} \right){{\left( {\ln \left( {n + 2} \right)} \right)}^3}}}$ converges.
Work Step by Step
To apply the Integral Test, we find the limits of integration by plotting the graph of the function $f\left( x \right) = \dfrac{1}{{\left( {x + 2} \right){{\left( {\ln \left( {x + 2} \right)} \right)}^3}}}$. From the figure attached, we see that $f$ is positive, decreasing, and continuous for $x \ge 0$. So, we evaluate the definite integral:
$\mathop \smallint \limits_0^\infty \dfrac{1}{{\left( {x + 2} \right){{\left( {\ln \left( {x + 2} \right)} \right)}^3}}}{\rm{d}}x$
Let $t = \ln \left( {x + 2} \right)$. So, $dt = \dfrac{1}{{x + 2}}dx$.
$\mathop \smallint \limits_0^\infty \dfrac{1}{{\left( {x + 2} \right){{\left( {\ln \left( {x + 2} \right)} \right)}^3}}}{\rm{d}}x = \mathop \smallint \limits_{\ln 2}^\infty \dfrac{1}{{{t^3}}}{\rm{d}}t = - \dfrac{1}{2}\left( {{t^{ - 2}}} \right)|_{\ln 2}^\infty $
$ = - \dfrac{1}{2}\left( { - {{\left( {\ln 2} \right)}^{ - 2}}} \right) = \dfrac{1}{{2{{\left( {\ln 2} \right)}^2}}} \approx 1.04068$
Since $\mathop \smallint \limits_0^\infty \dfrac{1}{{\left( {x + 2} \right){{\left( {\ln \left( {x + 2} \right)} \right)}^3}}}{\rm{d}}x$ converges, by the Integral Test, the series $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{\left( {n + 2} \right){{\left( {\ln \left( {n + 2} \right)} \right)}^3}}}$ also converges.
