Evaluating $\lim_{n\to \infty}\sum_{k=1}^{n}\frac{\sin{\frac{a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}$ (and another) 
Compute
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{\sin{\frac{a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}\right)$$ and
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{\sin{\frac{a}{3^k}}\sin{\frac{2a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}\right)$$
where $a$ is a positive real number.

I think the second limit may be solved by using the comparison criteria(limit comparison test), putting $-1 \le \sin{\frac{2a}{3^k}} \le 1$ and using the first limit. I suspect the first limit is $0$, so the second one should be $0$ too, but I might be wrong. I'm not sure how would you do the first one - I tried using the comparison test on the first one, but I didn't succeed. Can you help?

The second sum is telescoping, due to $$\frac{\sin x\sin 2x}{\sin 3x}=\frac14(\cot x-3\cot 3x).$$ This easily gives $$\sum_{k=1}^n\frac{\sin\frac{a}{3^k}\sin\frac{2a}{3^k}}{3^k\sin\frac{a}{3^{k-1}}}=\frac14\sum_{k=1}^n\left(\frac{1}{3^k}\cot\frac{a}{3^k}-\frac{1}{3^{k-1}}\cot\frac{a}{3^{k-1}}\right)=\frac14\left(\frac{1}{3^n}\cot\frac{a}{3^n}-\cot a\right).$$
The first sum doesn't seem to. Trying to replace $\cot$ by $-\csc$ yields$$\frac14(3\csc 3x-\csc x)=\frac{\sin^2 x}{\sin 3x},$$ so it would telescope with the sine in the numerator squared.
