# 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

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\mrm{f}\pars{x} = \lim_{n\to \infty}\sum_{r = 1}^{n}3^{r - 1}\sin^{3}\pars{x \over 3^{r}}:\ ?}$.