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

\begin{align}
\mrm{f}\pars{x} & =
\lim_{n\to \infty}\sum_{r = 1}^{n}3^{r - 1}\sin^{3}\pars{x \over 3^{r}} =
{1 \over 4}\lim_{n\to \infty}\sum_{r = 1}^{n}\bracks{%
3^{r}\sin\pars{x \over 3^{r}} - 3^{r - 1}\sin\pars{x \over 3^{r - 1}}}
\\[1cm] & =\require{cancel}
{1 \over 4}\lim_{n\to \infty}\left\lbrace%
\bracks{\cancel{3\sin\pars{x \over 3}} - \color{#f00}{\sin\pars{x}}} +
\bracks{\cancel{3^{2}\sin\pars{x \over 3^{2}}} - \cancel{3\sin\pars{x \over 3}}}\right. +
\\[5mm] &\phantom{= {1 \over 4}\lim_{n \to \infty}\braces{\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!}}
\left.\bracks{\cancel{3^{3}\sin\pars{x \over 3^{3}}} - \cancel{3^{2}\sin\pars{x \over 3^{2}}}} + \cdots +
\bracks{\color{#f00}{3^{n}\sin\pars{x \over 3^{n}}} -
\cancel{3^{n - 1}\sin\pars{x \over 3^{n - 1}}}}\!\!\right\rbrace
\\[1cm] & =
{1 \over 4}\,\lim_{n \to \infty}\bracks{-\sin\pars{x} +
3^{n}\sin\pars{x \over 3^{n}}} =
\bbx{{1 \over 4}\bracks{\vphantom{\Large a}x - \sin\pars{x}}}
\end{align}
