<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] 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:
$$ S = \frac{1}{2}\int_{0}^{1}\frac{7-8x+x^2}{1-x}\,dx = \frac{1}{2}\int_{0}^{1}(7-x)\,dx = \frac{1}{2}\left(7-\frac{1}{2}\right)=\color{red}{\frac{13}{4}}.\tag{3} $$
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