<s>[INST] Find the sum of the series $\sum_{n=1}^{\infty} \frac{2n+3}{n(n+1)(n+2)}$ 
Find the sum of the series $\sum_{n=1}^{\infty} \frac{2n+3}{n(n+1)(n+2)}$

My attempt: 
I tried partial fractions decomposition and I get :
$$\sum_{n=1}^{\infty} \frac{2n+3}{n(n+1)(n+2)}=\sum_{n=1}^{\infty}\frac {3}{2n}-\frac 1{n+1}-\frac 1{2(n+2)}=\frac{1}{2}\sum_{n=1}^{\infty}\frac 3{n}-\frac 2{n+1}-\frac 1{n+2}.$$
This should be the solution but the partial sum sequence I can't figure out the formula... what should I do in this case?
 [/INST] Good answers have already been given; here's some intuition:
$$\sum_{n=1}^{\infty}\frac{3}{n}-\frac{2}{n+1} - \frac{1}{n+2} = 3\left(1+\frac{1}2 +\frac{1}3 + \ldots\right) - 2\left(\frac{1}2 +\frac{1}3 + \ldots\right)-1\left(\frac{1}3 + \ldots\right)= 3+\frac{3}{2}+3\left(\frac{1}3 + \ldots\right) -1 -2\left(\frac{1}3 + \ldots\right) -1\left(\frac{1}3 + \ldots\right) = 3+\frac{1}{2}+3\left(\frac{1}3 + \ldots\right) -3 \left(\frac{1}3 + \ldots\right) = 3+\frac{1}{2}.$$
Now divide by two to get your answer.
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