Let's evaluate the ratio test:
[tex]\dfrac{a_n}{a_{n-1}}=\dfrac{3^{n+1}}{3n+1}\cdot\dfrac{3(n-1)+1}{3^n}=\dfrac{3\cdot3^n}{3n+1}\cdot\dfrac{3n-2}{3^n} = \dfrac{9n-6}{3n+1}[/tex]
The limit as [tex]n \to\infty[/tex] of this quantity is 3, which is more than 1. So, the series diverges.
