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$$\sum_{r=0}^∞ \frac {\sin(r θ)}{3^r} $$

Given: $\sinθ=1/3$

I recently came across this question, and I tried numerous ways of solving it, but I could reach nowhere. My initial approach was to expand uptil the first 3 terms to see if there was a pattern, but I don't think there is one. I also tried converting the question into a telescopic series, but I don't see how that can be done either. Any help would be appreciated!

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The sum is $$\Im\sum_{r=0}^\infty\left(\frac13\exp i\theta\right)^r=\Im\frac{1}{1-\frac13\exp i\theta}=\Im\frac{1-\frac13\exp -i\theta}{\frac{10}{9}-\frac23\cos\theta}=\frac{3\sin\theta}{10-6\cos\theta}.$$For $\sin\theta=\frac13$, $\cos\theta=\pm\frac{2\sqrt{2}}{3}$, so the sum simplifies to $\frac{1}{10\mp 4\sqrt{2}}=\frac{5\pm 2\sqrt{2}}{34}$.

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  • $\begingroup$ Still think that this is right, this is precisely what rolls out of WolframAlpha. $\endgroup$
    – RMWGNE96
    May 10 '19 at 12:01
  • $\begingroup$ any alternate method, mate? I haven't yet studied complex numbers $\endgroup$
    – ibuprofen
    May 10 '19 at 12:03
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    $\begingroup$ @MathDude3013 Ooh, that's a tough one. If you had spotted a pattern, you could prove it by induction. Spoiler alert: the sum of the first $n$ terms is $$\frac{\frac{1}{3}\sin\theta-\frac{1}{3^{n}}\sin n\theta+\frac{1}{3^{n+1}}\sin\left(n-1\right)\theta}{\frac{10}{9}-\frac{2}{3}\cos\theta}.$$As $n\to\infty$, only the numerator's first term survives. $\endgroup$
    – J.G.
    May 10 '19 at 12:09
  • $\begingroup$ well, that is very hard to spot :( $\endgroup$
    – ibuprofen
    May 10 '19 at 12:18
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    $\begingroup$ in about 4 months, I will be one of those people $\endgroup$
    – ibuprofen
    May 10 '19 at 12:21
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Use geometric series. We have $$\begin{split}\sum_{r=0}^\infty\frac{\sin(r\theta)}{3^r} &=\sum_{r=0}^\infty\mathrm{Im}\left(\frac{e^{ir\theta}}{3^r}\right) \\&=\mathrm{Im}\left(\sum_{r=0}^\infty\left(\frac{e^{i\theta}}{3}\right)^r\right)\\&=\mathrm{Im}\left(\frac{1}{1-e^{i\theta}/3}\right)\\&=\mathrm{Im}\left(\frac{1+e^{-i\theta}/3}{1+(e^{i\theta}-e^{-i\theta})/3-1/9}\right) \\&=\mathrm{Im}\left(\frac{1+\cos(\theta)/3-i\sin(\theta)/3}{8/9+2\sin(\theta)/3}\right)\\&=\frac{-1/9}{8/9+2/9}\\&=-\frac{1}{10}\end{split}$$ if I didn't make a silly mistake somewhere.

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  • $\begingroup$ Should the 3 in the exponent of the second line be an $r$? $\endgroup$
    – JMJ
    May 10 '19 at 11:56
  • $\begingroup$ Yes, edited accordingly $\endgroup$
    – RMWGNE96
    May 10 '19 at 11:59
  • $\begingroup$ You have some sign errors, which is why you got a sine, not a cosine, in your denominator. (The $1/9$ has a sign error too, but they're linked.) $\endgroup$
    – J.G.
    May 10 '19 at 12:22
  • $\begingroup$ Yes, you're right! You might want to edit my answer $\endgroup$
    – RMWGNE96
    May 10 '19 at 12:26

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