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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.

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  • $\begingroup$ Yes. If you would like I could edit my question to make it a little bit clearer $\endgroup$
    – Aditya Sriram
    Commented Jan 24, 2020 at 11:29

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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

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  • $\begingroup$ Thank you for your answer. However I would like to calculate the actual sum. I figured out that the series will converge. $\endgroup$
    – Aditya Sriram
    Commented Jan 24, 2020 at 11:21

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