<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] $$\frac{3n+7}{n(n+1)(n+2)}=\frac{3n+6+1}{n(n+1)(n+2)}=$$
$$=\frac{3}{n(n+1)}+\frac{1}{n(n+1)(n+2)}=\frac{3}{n}-\frac{3}{n+1}+\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=$$
$$=\frac{3}{n}-\frac{3}{n+1}+\frac{1}{n}-\frac{1}{n+1}-\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)=$$
$$=\frac{7}{2n}-\frac{4}{n+1}+\frac{1}{2(n+2)}$$ and use the telescopic sum
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