# 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?

• The terms of the decomposition you've got all diverge Nov 14, 2018 at 18:54
• You should use $2n+3=(n+1)+(n+2)$ Nov 14, 2018 at 18:55
• @YuriyS I know but what then? Nov 14, 2018 at 18:58
• You get the same thing Poon Levi shows in the answer Nov 14, 2018 at 19:01
• Mathematica gives $7/4$. Nov 14, 2018 at 19:07

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.