# Sum of series $\sum \limits_{k=1}^{\infty}\frac{\sin^3 3^k}{3^k}$

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?

• 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? – Mark Fischler Jan 23 '17 at 7:20
• 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. – ZFR Jan 23 '17 at 7:24
• @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. – epimorphic Jan 25 '17 at 3:31
• @RFZ: You didn't sign the right answer (until now) ? – Khosrotash Sep 21 '17 at 7:42
• @Khosrotash, You want likes? :) Oh man this sounds awful :) – ZFR Sep 21 '17 at 10:40

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 $$\sum_{k\geq1}\frac{\sin^{3}\left(3^{k}\right)}{3^{k}}=\frac{1}{4}\sum_{k\geq 0}\frac{\left(-1\right)^{k}}{\left(2k+1\right)!}3^{2k+1}=\color{red}{\frac{\sin\left(3\right)}{4}}.$$

• badass $(+1)$ – tired Jan 23 '17 at 13:09

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}

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 $$\frac {\sin (3)} {4}$$.

• +1 Pretty fine. The obvious way. – Felix Marin Sep 11 '18 at 20:33