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?
 A: 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 $\color{red}{\frac{13}{4}}$.
A: $$\frac{3n+7}{n(n+1)(n+2)}=\frac{3n+6+1}{n(n+1)(n+2)}=$$
$$=\frac{3}{n(n+1)}+\frac{1}{n(n+1)(n+2)}=\frac{3}{n}-\frac{3}{n+1}+\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=$$
$$=\frac{3}{n}-\frac{3}{n+1}+\frac{1}{n}-\frac{1}{n+1}-\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)=$$
$$=\frac{7}{2n}-\frac{4}{n+1}+\frac{1}{2(n+2)}$$ and use the telescopic sum
A: $S=\sum_{n=1}^{+\infty}(\frac {3}{(n+1)(n+2)}+\frac {7}{n (n+1)(n+2)}) $
$$=3\sum_{n=1}^{+\infty}(\frac {1}{n+1}-\frac {1}{n+2}) $$
$$+7\sum_{n=1}^{+\infty}\frac{1}{n (n+1)(n+2)} $$

$$\implies S=\frac {3}{2}+7A$$

and
$$S=\sum_{n=1}^{+\infty}\frac {3 (n+2)+1}{n (n+1)(n+2)} $$
$$=3\sum_{n=1}^{+\infty}(\frac {1}{n}-\frac {1}{n+1})+A$$

$$\implies S=3+A $$

thus after eliminating $A $, we get
$$6S=21-\frac {3}{2}\implies S=\frac {13}{4} .$$
A: Alternatively:
$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}=\sum_{n=1}^{\infty}\left(\frac{4}{n(n+1)}-\frac{1}{n(n+2)}\right)=$$
$$4\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac 12\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right)=$$
$$4\left(1-\require{cancel}\cancel{\frac 12}+\cancel{\frac12}-\cancel{\frac13}+\cancel{\frac13}-\cancel{\frac14}+...\right)-\frac12\left(1-\cancel{\frac13}+\frac12-\cancel{\frac14}+\cancel{\frac13}-\cancel{\frac15}+...\right)=$$
$$4-\frac{1}{2}\cdot\frac32=\frac{13}{4}.$$
A: As an alternative approach, since by computing residues we have
$$ \frac{3n+7}{n(n+1)(n+2)} = \frac{7}{2}\cdot\frac{1}{n}-4\cdot\frac{1}{n+1}+\frac{1}{2}\cdot\frac{1}{n+2} \tag{1}$$
it follows that
$$ S=\sum_{n\geq 1}\frac{3n+7}{n(n+1)(n+2)} = \int_{0}^{1}\left(\frac{7}{2}-4x+\frac{x^2}{2}\right)\sum_{n\geq 1}x^{n-1}\,dx \tag{2}$$
hence:
$$ S = \frac{1}{2}\int_{0}^{1}\frac{7-8x+x^2}{1-x}\,dx = \frac{1}{2}\int_{0}^{1}(7-x)\,dx = \frac{1}{2}\left(7-\frac{1}{2}\right)=\color{red}{\frac{13}{4}}.\tag{3} $$
