$$f(x) = \lim_{n\to \infty} \sum_{r=1}^n 3^{r-1}\sin^3(x/(3^r)) $$ He intentado utilizar la fórmula relativa $\sin(3x)$$\sin^3(x)$, pero recibió más tarde pegado con una serie similar que la suma no sabía cómo calcular
Respuesta
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove armada]{{#1}}\,} \newcommand{\llaves}[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}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\mrm{f}\pars{x} = \lim_{n\to \infty}\sum_{i = 1}^{n}3^{i - 1}\sin^{3}\pars{x \a más de 3^{r}}:\ ?}$.
\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}
