# Limit of the given sum: $f(x) = \lim_{n\to \infty} \sum_{r=1}^n 3^{r-1}\sin^3(x/(3^r))$

$$f(x) = \lim_{n\to \infty} \sum_{r=1}^n 3^{r-1}\sin^3(x/(3^r))$$ I tried using the formula relating $\sin(3x)$ to $\sin^3(x)$ but got later stuck with a similar series who's sum I didn't know how to calculate

• I don't understand what is being summed here. Is the equation as stated correct? Nov 4, 2016 at 20:52
• I'm trying to but I can't get to place the infinity sign i n the limit Nov 4, 2016 at 20:55
• I've edited it to try and make sense of your formula but you might want to check if this is actually what you intended to ask. Nov 4, 2016 at 20:56
• I feel like there is a good chance that this always diverges because $3^{n-1}$ grows exponentially, while the sum is bounded and does not, to my knowledge, converge to zero. Nov 4, 2016 at 21:03
• Note that $$\sin^3\theta=\dfrac{1}{4}(3\sin\theta-\sin 3\theta)$$ and $$\sum_{r=0}^n\sin k\theta=\dfrac{\sin\left(\dfrac{n\theta}{2}\right)\sin \left(\dfrac{(n-1)\theta}{2}\right)}{\sin \left(\dfrac{\theta}{2}\right)}$$ Nov 4, 2016 at 21:47

$\ds{\mrm{f}\pars{x} = \lim_{n\to \infty}\sum_{r = 1}^{n}3^{r - 1}\sin^{3}\pars{x \over 3^{r}}:\ ?}$.