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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?
Hint: The last sum can be written as $$\sum_{n=1}^\infty\left[2\left(\frac1n-\frac1{n+1}\right)+\left(\frac1n-\frac1{n+2}\right)\right]$$ The is a telescoping sum and we can expicitly write down the sequence of partial sums.
One way is writing $$\sum_{n=1}^{\infty} \frac{n+1+n+2}{n(n+1)(n+2)}=\sum_{n=1}^{\infty} \frac{1}{n(n+2)}+\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$$ both are telescopic.
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.
