Question

Please and thankyou (:

Answer 1

Let's evaluate the ratio test:

$\dfrac{a_n}{a_{n-1}}=\dfrac{3^{n+1}}{3n+1}\cdot\dfrac{3(n-1)+1}{3^n}=\dfrac{3\cdot3^n}{3n+1}\cdot\dfrac{3n-2}{3^n} = \dfrac{9n-6}{3n+1}$

The limit as $n \to\infty$ of this quantity is 3, which is more than 1. So, the series diverges.