Calculate $\lim\limits_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{6}{k(k+1)(k+3)}$ $$\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{6}{k(k+1)(k+3)}$$
I tried to simplify the sum and I got $\frac{2}{k}-\frac{3}{k+1}+\frac{1}{k+3}$ but  I can't use this to simplify the terms.Also,I tried to amplify with $k+2$ and I got $$\frac{(k+3)-1}{k(k+1)(k+2)(k+3)}=\frac{(k+3)}{k(k+1)(k+2)(k+3)}-\frac{1}{k(k+1)(k+2)(k+3)}$$ but the terms also don't simplify.
 A: Hint
$$\frac{2}{k}-\frac{3}{k+1}+\frac{1}{k+3}=\frac{2}{k}-\frac{2}{k+1}-\frac{1}{k+1}+\frac{1}{k+3}=2\left(\frac{1}{k}-\frac{1}{k+1} \right)-\left(\frac{1}{k+1}-\frac{1}{k+3} \right)$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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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}}}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 1}^{\infty}{6 \over k\pars{k + 1}\pars{k+ 3}}} =
6\sum_{k = 1}^{\infty}2\int_{0}^{1}\dd u_{1}\int_{0}^{u_{1}}\dd u_{2}\,
{1 \over \pars{k + u_{1} + 2u_{2}}^{3}}
\\[5mm] = &\
12\int_{0}^{1}\dd u_{1}\int_{0}^{u_{1}}\dd u_{2}\,\
\underbrace{\sum_{k = 1}^{\infty}{1 \over \pars{k + u_{1} + 2u_{2}}^{3}}}
_{\ds{-\Psi''\pars{1 + u_{1} + 2u_{2}}/2}}\qquad
\pars{~\Psi:\ Digamma\ Function~}
\\[5mm] = &\
-3\int_{0}^{1}\dd u_{1}\bracks{\Psi\, '\pars{1 + 3u_{1}} -
\Psi\, '\pars{1 + u_{1}}}
\\[5mm] = &\
-3\bracks{{1 \over 3}\,\Psi\pars{1 + 3u_{1}} - \Psi\pars{1 + u_{1}}}_{\ 0}^{ 1} =
-\
\Psi\pars{4} + 3\Psi\pars{2} + \Psi\pars{1} - 3\Psi\pars{1}
\\[5mm] = &\
3\ \underbrace{\bracks{\Psi\pars{2} - \Psi\pars{1}}}_{\ds{=\ 1}}\ -\
\underbrace{\bracks{\Psi\pars{4} - \Psi\pars{1}}}
_{\ds{=\ {11 \over 6}}}\ =\ \bbx{7 \over 6}
\end{align}

Note that $\quad\left\{\begin{array}{rclcl}
\ds{\Psi\pars{2}} & \ds{=} & \ds{\Psi\pars{1} + {1 \over 1}} &&
\\[1mm]
\ds{\Psi\pars{3}} & \ds{=} & \ds{\Psi\pars{2} + {1 \over 2}}
& \ds{=} & \ds{\Psi\pars{1} + {3 \over 2}}
\\[1mm]
\ds{\Psi\pars{4}} & \ds{=} & \ds{\Psi\pars{3} + {1 \over 3}}
& \ds{=} & \ds{\Psi\pars{1} + {11 \over 6}}
\end{array}\right.$

