Does the series $\sum 2^n \sin(\frac{\pi}{3^n})$ converge? 
Check if $$\sum_{n = 1}^{\infty}2^n \sin\left(\frac{\pi}{3^n}\right)$$ converges.

I tried to solve this by using the ratio  test - I have ended up with the following limit to evaluate: 
$$\lim_{n \to \infty} \left(\frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)} \right)$$
And now - I am stuck and don't know how to proceed with this limit. Any hints?

If you want to stick with your approach using the ratio test, you can (although Olivier's answer is more direct). Caveat: I'll detail every step of the derivation, which is not actually necessary for a proof.
We will only rely on elementary arguments, specifically the fact that $\sin'(0)=\cos 0 = 1$ — which is equivalent to $$\lim_{x\to 0} \frac{\sin x}{x} = 1 \,.\tag{1}
$$
From there, you can write
$$
\lim_{n \to \infty} \frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)}
= \lim_{n \to \infty} 2\cdot \frac{1}{3}\cdot \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot\frac{\frac{\pi}{3^n}}{\sin \left(\frac{\pi}{3^n} \right)}
= \frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot\left(\frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1} \tag{2}
$$
and, by (1), we get
$$\begin{align*}
\lim_{n \to \infty} \frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)}
&=
\frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot \lim_{n \to \infty} \left(\frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1}
=\frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot \left(\lim_{n \to \infty} \frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1}  \\
&= \frac{2}{3}\cdot 1\cdot 1^{-1} = \boxed{\frac{2}{3}}
\end{align*}$$
and you can conclude with the ratio test.
