Answer
Divergent
Work Step by Step
Each term of the series has the form $\frac{(2)(5)(8)...(3n - 1)}{(3)(5)(7)...(2n + 1)}$
So, |$\frac{a_{n+1}}{a_{n}}$| = |$\frac{(2)(5)(8)...(3n - 1)(3n+2)}{(3)(5)(7)...(2n + 1)(2n + 3)}$| $\times$ |$\frac{(3)(5)(7)...(2n + 1)}{(2)(5)(8)...(3n - 1)}$| = $\frac{3n + 2}{2n + 3}$
Now, $\lim\limits_{n \to \infty}$$\frac{3n + 2}{2n + 3}$ = $\frac{3}{2}$ $\gt$ 1 which makes it divergent.
