¿cuál es la prueba para $\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} = \frac{1}{4} $

Question

¿Puede alguien proporcionar una prueba para la solución de esta serie

$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} = \frac{1}{4} $

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}{1 \over n\pars{n + 1}\pars{n + 2}} = {1 \over 4}:\ {\large ?}}$ \begin{align} {1 \over n\pars{n + 1}\pars{n + 2}}= {1/2 \over n} + {-1 \over n + 1} + {1/2 \over n + 2} =\half\pars{{1 \over n} - {1 \over n + 1}} - \half\pars{{1 \over n + 1} - {1 \over n + 2}} \end{align} Ahora, tienes series "telescópicas": \begin{align} &\color{#00f}{\large\sum_{n = 1}^{\infty}{1 \over n\pars{n + 1}\pars{n + 2}}} =\half\sum_{n = \color{#f00}{\large 1}}^{\infty}\pars{{1 \over n} - {1 \over n + 1}} -\half\sum_{n = \color{#f00}{\large 2}}^{\infty}\pars{{1 \over n} - {1 \over n + 1}} \\[3mm]&=\half\pars{{1 \over 1} - {1 \over 1 + 1}} =\color{#00f}{\large{1 \over 4}} \end{align}

Answer 2

$$\dfrac{1}{n(n+1)(n+2)}=\dfrac{1}{2}\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right)$$ así que \begin{align*}\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(n+2)}&=\dfrac{1}{2}\sum_{n=1}^{\infty}\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right)\\ &=\dfrac{1}{2}\lim_{n\to\infty}\left(\dfrac{1}{1\times 2}-\dfrac{1}{(n+1)(n+2)}\right)\\ &=\dfrac{1}{4} \end{align*}