Test $\sum_{n=1}^\infty [n^4 \sin^2 (\frac{2n}{3n^3 - 2n^2 + 5})]^n$ for convergence Any hints regarding this question would be appreciated - is the ratio test the best place to start?

Hint: 
$$a_n^{1/n} = n^4 \sin^2 \bigg( \frac{2n}{3n^3-2n^2+5} \bigg) \sim n^4 \bigg(\frac{2n}{3n^3-2n^2+5} \bigg)^2 = \frac{4n^6}{(3n^3-2n^2+5)^2}$$
