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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?

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

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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.

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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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