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?

Solution 1:

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}

Solution 2:

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

Solution 3:

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