Answer
Diverges.
Work Step by Step
Write $\dfrac{{k\left( {k + 3} \right)}}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 5} \right)}} = \dfrac{{{k^2} + 3k}}{{{k^3} + 8{k^2} + 17k + 10}}$. Notice that the leading term is $\dfrac{{{k^2}}}{{{k^3}}} = \dfrac{1}{k}$. Thus, we try to use $\dfrac{1}{k}$ in the Limit Comparison Test.
Let ${a_k} = \dfrac{{k\left( {k + 3} \right)}}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 5} \right)}}$ and ${b_k} = \dfrac{1}{k}$.
Evaluate:
$\rho = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{a_k}}}{{{b_k}}} = \dfrac{{k\left( {k + 3} \right)}}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 5} \right)}}\cdot k = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{k^3} + 3{k^2}}}{{{k^3} + 8{k^2} + 17k + 10}}$
$ = \mathop {\lim }\limits_{k \to \infty } \dfrac{{1 + \dfrac{3}{k}}}{{1 + \dfrac{8}{k} + \dfrac{{17}}{{{k^2}}} + \dfrac{{10}}{{{k^3}}}}} = 1$
Since $\rho = 1 \gt 0$ and the harmonic series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{k}$ diverges, by the Limit Comparison Test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{{k\left( {k + 3} \right)}}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 5} \right)}}$ diverges.
