<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] $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
$$6S=21-\frac {3}{2}\implies S=\frac {13}{4} .$$
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