I came across this summation problem the other day and I am not quite sure how to approach it
$$S=\sum_{n=0}^{n=\infty}\frac{2^{n-1}}{3^{2n-2}}\sin\left(\frac{\pi}{3.2^{n-1}}\right)$$
My approach involved complex numbers where I assumed $$z=\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}$$ and so the sum reduced to $$\frac{4S}{3}=\sum_{n=0}^{n=\infty}\left(\frac{2}{9}\right)^{n}z^{\frac{1}{2^n}}$$ However I am not able to reduce this any further because it is neither an AP nor a GP.
Help would be much appreciated. Thanks
P.S - I am expected to find a closed-form solution to the above sum.
Ratio test:
$$R=\frac{\bigg(\frac29\bigg)^{n+1}z^{\frac{1}{2^{n+2}}}} {\bigg(\frac29\bigg)^{n} z^{\frac{1}{2^n}}}=\frac 29 z^{\frac 14}$$
$|R|<1$ implies convergence