<s>[INST] Sum $\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$ How can we find the sum 
$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$$
WolframAlpha says it's $13/4$ but how to compute it? In general how to approach this kind of sums of rational functions where denominator has distinct roots?
 [/INST] Using partial fractions we have
\begin{eqnarray*}
 \frac{3n+7}{n(n+1)(n+2)}= \frac{\frac{7}{2}}{n}+\frac{-4}{n+1}+\frac{\frac{1}{2}}{n+2}
\end{eqnarray*}
It is a telescoping sum ...
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{3n+7}{n(n+1)(n+2)} = \frac{\frac{7}{2}}{1}&+&\color{green}{\frac{-4}{2}} &+&\color{red}{\frac{\frac{1}{2}}{3}} \\&+&\color{green}{\frac{\frac{7}{2}}{2}}&+&\color{red}{\frac{-4}{3}+\frac{\frac{1}{2}}{4}} \\
&& &+&\color{red}{\frac{\frac{7}{2}}{3}}+\color{red}{\frac{-4}{4}+\frac{\frac{1}{2}}{5}} \\
 &&&& \ddots
\end{eqnarray*} 
After the dust settles ... we have $\color{red}{\frac{13}{4}}$.
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