Show that series $$\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}$$ converges by simplifying its sequence of partial sums and find its sum.
I don't have much detail but this all I have:
$$\begin{align}\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)} &=\lim_{m\to\infty}\sum_{n=1}^{m}\frac{6}{m(m+1)(m+2)}\\ &=\lim_{m\to\infty}\frac{3(m^2+3m)}{2(m+1)(m+2)}\\ &=\frac{3}{2}\\ \end{align}$$
I know, I don't have much but any help will do with the detail or is it right.
On
$$\begin{align} \sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)} &=\sum_{n=1}^{\infty}\left(\frac{3}{n}+\frac{3}{n+2}-\frac{6}{n+1}\right)\tag{1}\\ &=\sum_{n=1}^{\infty}\int_{0}^{1}\left(3x^{n-1}+3x^{n+1}-{6x}^{n}\right)\,\mathrm dx\tag{2}\\ &=\int_{0}^{1}\sum_{n=1}^{\infty}\left(3x^{n-1}+3x^{n+1}-{6x}^{n}\right)\,\mathrm dx\tag{3}\\ &=\int_{0}^{1}\frac{3}{1-x}+\frac{3x^2}{1-x}-\frac{6x}{1-x}\,\mathrm dx\tag{4}\\ &=\int_{0}^{1}\left(\frac{3+3x^2-6x}{1-x}\right)\,\mathrm dx\tag{5}\\ &=3\int_{0}^{1}\frac{x^2-2x+1}{1-x}\,\mathrm dx\tag{6}\\ &=3\int_{0}^{1}{1-x}\,\mathrm dx\tag{7}\\ &=3\cdot\frac14\\ &=\frac32\\ \end{align}$$
$$\large\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}=\frac32\\$$
On
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \sum_{n\ =\ 1}^{\infty}x^{n - 1}={1 \over 1 -x}\ \imp\ \sum_{n\ =\ 1}^{\infty}{x^{n} \over n}=\int_{0}^{x}{\dd t \over 1 - t} =-\ln\pars{1 -x} $$
$$ \sum_{n\ =\ 1}^{\infty}{x^{n + 1} \over n\pars{n + 1}} =-\int_{0}^{x}\ln\pars{1 - t}\,\dd t =x + \pars{1 - x}\ln\pars{1 - x} $$
$$ \sum_{n\ =\ 1}^{\infty}{1 \over n\pars{n + 1}\pars{n + 2}} =\int_{0}^{1}\bracks{t + \pars{1 - t}\ln\pars{1 - t}}\,\dd t ={1 \over 4} $$
$$\color{#66f}{\large% \sum_{n\ =\ 1}^{\infty}{6 \over n\pars{n + 1}\pars{n + 2}}} =6\times{1 \over 4}=\color{#66f}{\large{3 \over 2}} $$
Since \begin{gather*} \frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right), \end{gather*} we have, by telescoping summation, \begin{align*} \sum_{n=1}^N\frac{1}{n(n+1)(n+2)}&=\frac{1}{2}\sum_{n=1}^N\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\\ &=\frac{1}{2}\sum_{n=1}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{n=1}^N\frac{1}{(n+1)(n+2)}\\ &=\frac{1}{2}\sum_{n=1}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{m=2}^{N+1}\frac{1}{m(m+1)}\quad \quad (m=n+1)\\ &=\frac{1}{2}\sum_{n=1}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{n=2}^{N+1}\frac{1}{n(n+1)}\\ &=\frac{1}{2}\cdot\frac{1}{2}+\sum_{n=2}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{n=2}^{N}\frac{1}{n(n+1)}-\frac{1}{2}\cdot\frac{1}{(N+1)(N+2)}\\ &=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{(N+1)(N+2)}\right)\to \frac{1}{4}, \qquad \text{as } N\to\infty, \end{align*} from which we can infer that \begin{gather*} \sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}=6\cdot\frac{1}{4}=3/2. \end{gather*}