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.

For $n\geq 1$ we have $3^n>3^n-2>0$ so $0<3^{-n}<(3^n-2)^{-1}$ so $4^{n+1}(3^n-2)^{-1}>4^{n+1}3^{-n}=4 (4/3)^n>4.$ If the terms of the series do not converge to $0$ then the sum cannot be convergent.
