Prove the convergence and find the sum of series $\sum\limits_{n=1}^{\infty}\left(n^3\sin\frac{\pi}{3^n}\right)$. We know that $0<\sin\frac{\pi}{3^n}\le\frac{\sqrt 3}{2},\forall n\ge 1$. How to find the boundary for $n^3\sin\frac{\pi}{3^n}$ (how to use comparison test here)?
I tried using the ratio test, but the limit $$\lim\limits_{n\to\infty}\left((n+1)^3\sin\frac{\pi}{3^{n+1}}\cdot \frac{1}{n^3\sin\frac{\pi}{3^n}}\right)$$ isn't that easy to evaluate.

You can evaluate the limit in this way
$$\begin{align}
\lim_{n\to\infty} \left( \frac{(n+1)^3}{n^3} \cdot \frac{\sin\frac{\pi}{3^{n+1}}}{\sin\frac{\pi}{3^{n}}}\right) &= \lim_{n \to \infty} \frac{(n+1)^3}{n^3} \lim_{n \to \infty} \frac{\sin\frac{\pi}{3^{n+1}}}{\sin\frac{\pi}{3^{n}}} \\
&= 1 \cdot \lim_{n \to \infty} \frac{\frac{\pi}{3^{n+1}}}{\frac{\pi}{3^{n}}} \\
&= 1 \cdot \frac{1}{3} \\
&= \frac{1}{3}
\end{align}$$
and since the result is less than $1$ the series will converge.
