<s>[INST] 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.
 [/INST] Since $\sin x \le x$ for $x\ge 0$ and $0\le \frac{\pi}{3^n}\le \pi$,
$$
0\le n^3 \sin\frac{\pi}{3^n} \le \frac{n^3 \pi}{3^n}
$$
and $\sum_{n=1}^{\infty}\frac{n^3 \pi}{3^n}$ converges by ratio test. By comparison test, given series converges.
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