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Calculate the following sum: $$\sum \limits_{k=1}^{\infty}\dfrac{\sin^3 3^k}{3^k}$$

Unfortunately I have no idea how to handle with this problem.

Could anyone show it solution?

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    $\begingroup$ You probably have a reason to hope this can be done in closed form (perhaps it was a contest problem, for example). Otherise, I would never think it could be done. What, may I ask, is your reason for optimism? $\endgroup$ Jan 23, 2017 at 7:20
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    $\begingroup$ I am not sure is it a contest problem or not, but i found in one of the papers of problem-solving seminars of my university. $\endgroup$
    – ZFR
    Jan 23, 2017 at 7:24
  • $\begingroup$ @FelixMarin I believe I've warned you not to run amok introducing personal stylistic quirks to other people's posts. \frac is the proper way to write fractions, not \over. Question marks are not to be preceded with a space in English, and it's downright ridiculous to add a period afterwards. $\endgroup$
    – epimorphic
    Jan 25, 2017 at 3:31
  • $\begingroup$ @RFZ: You didn't sign the right answer (until now) ? $\endgroup$
    – Khosrotash
    Sep 21, 2017 at 7:42
  • $\begingroup$ @Khosrotash, You want likes? :) Oh man this sounds awful :) $\endgroup$
    – ZFR
    Sep 21, 2017 at 10:40

3 Answers 3

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Using $$\sin(3a)=3\sin a-4\sin^3a \to \color{red}{\sin^3(a)=\frac14\Big(3\sin a-\sin(3a)\Big)} $$

so \begin{eqnarray} \sum_{k=1}^{\infty}\frac{\sin^3(3^k)}{3^k} &=& \frac14\sum_{k=1}^{\infty}\frac{3\sin(3^k)-\sin(3.3^k)}{3^k}\\ &=& \frac14\sum_{k=1}^{\infty}\frac{\sin(3^k)}{3^{k-1}}-\frac{\sin(3^{k+1})}{3^{k}}\\ &=& \frac14\sum_{k=1}^{\infty}f(k)-f(k+1)\\ &=&\frac14\Big(\frac{\sin3}{3^{1-1}}-\lim_{n \to \infty}\frac{\sin(3^{n+1})}{3^n}\Big)\\ &=&\frac{\sin(3)}{4} \end{eqnarray}

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    $\begingroup$ Pov: You unlocked god level. (+1) $\endgroup$ Sep 17, 2021 at 18:30
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    $\begingroup$ @LaxmiNarayanBhandari: thank you a ton. $\endgroup$
    – Khosrotash
    Oct 1, 2021 at 7:49
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    $\begingroup$ I wonder why your answer didn't get accepted. It is literally the best way to do it. $\endgroup$ Oct 1, 2021 at 11:57
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An overkill. Let $\mathfrak{M}\left(*,s\right) $ the Mellin transform. Using the identity $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\lambda_{k}g\left(\mu_{k}x\right),\, s\right)=\underset{k\geq1}{\sum}\frac{\lambda_{k}}{\mu_{k}^{s}}\mathfrak{M}\left(g\left(x\right),s\right) $$ we have $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\frac{\sin^{3}\left(3^{k}x\right)}{3^{k}},\, s\right)=\underset{k\geq1}{\sum}\left(\frac{1}{3^{s+1}}\right)^{k}\mathfrak{M}\left(\sin^{3}\left(x\right),s\right) $$ and since $$\mathfrak{M}\left(\sin^{3}\left(x\right),s\right)=\frac{\Gamma\left(s\right)\sin\left(\frac{\pi}{2}s\right)}{4}\left(3-\frac{1}{3^{s}}\right) $$ we have for $\textrm{Re}\left(s\right)>-1 $ $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\frac{\sin^{3}\left(3^{k}x\right)}{3^{k}},\, s\right)=\frac{\Gamma\left(s\right)\sin\left(\frac{\pi}{2}s\right)}{4}\left(3-\frac{1}{3^{s}}\right)\frac{1}{3^{s+1}-1}=\frac{\Gamma\left(s\right)\sin\left(\frac{\pi}{2}s\right)3^{-s}}{4} $$ and so inverting we get $$\underset{k\geq1}{\sum}\frac{\sin^{3}\left(3^{k}x\right)}{3^{k}}=\frac{1}{2\pi i}\int_{1/2-i\infty}^{1/2+i\infty}\frac{\Gamma\left(s\right)\sin\left(\frac{\pi}{2}s\right)3^{-s}}{4}x^{-s}ds $$ now taking $x=1$ and shifting the complex line to the left we have, from the residue theorem, that we have to evaluate the residues of $$\textrm{Res}_{s=-2n-1} \frac{\Gamma\left(s\right)\sin\left(\frac{\pi}{2}s\right)3^{-s}}{4}=\frac{\left(-1\right)^{k}}{4\left(2k+1\right)!3^{-2n-1}} $$ hence $$\sum_{k\geq1}\frac{\sin^{3}\left(3^{k}\right)}{3^{k}}=\frac{1}{4}\sum_{k\geq 0}\frac{\left(-1\right)^{k}}{\left(2k+1\right)!}3^{2k+1}=\color{red}{\frac{\sin\left(3\right)}{4}}.$$

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    $\begingroup$ badass $ (+1) $ $\endgroup$
    – tired
    Jan 23, 2017 at 13:09
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We can use the following identity (leaving for you to prove)

$$ \sin^3 (3^x) = \frac {1}{4} ( 3 \sin(3^x) - \sin (3^{x+1}) ) $$

That is,

$$ \frac {\sin^3( 3) }{3} = \frac {1}{4} ( \sin (3) - \frac {1}{3} \sin (3^2) ) $$ $$ \frac {\sin^3( 3^2) }{3^2} = \frac {1}{4} (\frac {1}{3} \sin (3^2) - \frac {1}{3^2} \sin (3^3) ) $$ $$ \frac {\sin^3( 3^3) }{3^3} = \frac {1}{4} (\frac {1}{3^3} \sin (3^3) - \frac {1}{3^4} \sin (3^4) ) $$ As you can clearly see, all the middle terms cancel out in the sum and we are only left with the first term. So the sum is $$\frac {\sin (3)} {4} $$.

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  • $\begingroup$ +1 Pretty fine. The obvious way. $\endgroup$ Sep 11, 2018 at 20:33

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