## University Calculus: Early Transcendentals (3rd Edition)

Consider $a_n=\dfrac{n! \ln n}{n(n+2)!}$ Thus, $\Sigma_{n=1}^\infty \dfrac{n! \ln n}{n(n+2)!} \leq \Sigma_{n=1}^\infty \dfrac{n}{n(n+1)(n+2)}\leq \Sigma_{n=1}^\infty \dfrac{n}{n^3}=\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ we can see that the series $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ is a convergent p-series. Hence, the given series converges by the direct comparison test.