Prove that : $\sum n^{3}\sin\frac{1}{3^{n}}$ converges.

Question

I'm thinking of applying the ratio test. That would give me $\lim\limits_{n \to \infty } \dfrac {(n+1)^3}{n^3} \dfrac {\sin\frac {1}{3^{n+1}}}{\sin \frac {1}{3^n}}$.

The first part is $1$ as $n$ approaches infinity, but what about the sine part? Or am I not doing this correctly?

Answer 1

hint: $\sin x < x , x = 3^{-n}$, then use the fact that $\dfrac{n^3}{3^n} \to 0$

Answer 2

For large $n$, we can taylor expand $\sin(1/3^n)\sim 1/3^n$, which is enough to note that $$ \sum n^3\sin 1/3^n\sim \sum n^3/3^n $$ Which converges.