Decide if the series $\sum\frac{4^{n+1}}{3^{n}-2}$ converges or diverges and, if it converges, find its sum 
Decide if the series $$\sum_{n=1}^\infty\frac{4^{n+1}}{3^{n}-2}$$ converges or diverges and, if it converges, find its sum.

Is this how you would show divergence attempt:
For $n \in [1,\infty), a_n = \frac{4^{n+1}}{3^n -2} \geq 0$
For $n \in [1,\infty), a_n = \frac{4^{n+1}}{3^n-2} \geq \frac{4^{n+1}}{3^n} = b_n$
Since $\sum_{n=1}^{\infty} \frac{4^{n+1}}{3^n}$ is a geometric series with $r = \frac{4}{3} > 1$. Therefore it diverges by the geometric series test and by the comparison test $\sum a_n$ diverges  too.

The base sequence is not an infinitesimal (necessary condition of convergence). Automatically, the series diverges.
