Answer
The series is absolutely convergent.
Work Step by Step
$a_{n}=\Sigma_{n=1}^{ \infty}(-1)^{n}\frac{2^{n}n!}{5.8.11.......(3n+2)}$
$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(-1)^{n+1}\frac{2^{n+1}(n+1)!}{5.8.11.......(3n+2)(3n+5)}}{(-1)^{n}\frac{2^{n}n!}{5.8.11.......(3n+2)}}|$
$=\lim\limits_{n \to \infty}\frac{2(n+1)}{3n+5}$
$=\frac{2}{3}\lt 1$
The series is absolutely convergent.