Another Telescoping Series

Let $S_N=\displaystyle\sum_{m=0}^{N} \dfrac{\sin^{3}({3^m})}{3^m}$

$\rightarrow S_N=\dfrac{3}{4}\displaystyle\sum_{m=0}^{N} \dfrac{4\sin^{3}({3^m})}{3^{m+1}}$

As we know that $\sin (3x)=3\sin{x}-4\sin^{3}{x}$

So $\rightarrow 4\sin^{3}{x}=3\sin{x}-\sin (3x)$

$\rightarrow S_N=\dfrac{3}{4}\displaystyle\sum_{m=0}^{N} \left(\dfrac{\sin {(3^m)}}{3^{m}}-\dfrac{\sin {(3^{m+1})}}{3^{m+1}}\right)$

Now, Let $\dfrac{\sin {(3^m)}}{3^{m}}=A_m$

So, $\dfrac{\sin {(3^{m+1})}}{3^{m+1}}=A_{m+1}$

$\rightarrow S_N=\dfrac{3}{4}\displaystyle\sum_{m=0}^{N} A_m-A_{m+1}$

$\rightarrow S_N=\dfrac{3}{4} \left[(A_0-A_1)+(A_1-A_2)+...........+(A_{N-1}-A_{N})+(A_{N}-A_{N+1})\right]$

$\rightarrow S_N=\dfrac{3}{4}(A_0-A_{N+1})$

$\rightarrow S_N=\dfrac{3}{4} \left(\sin 1-\dfrac{\sin (3^{N+1})}{3^{N+1}}\right)$

Finally, as $N \rightarrow \infty$ and beyond, :D

$S_{\infty}=\dfrac{3\sin {1}}{4} \approx \boxed{0.6311}$