I need to calculate this sum. what i tried is doing partial fraction decomposition.
so $$\frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}$$
and after calculating I got $$A = \frac{1}{2}$$ $$B = -\frac{1}{4}$$ $$C = \frac{1}{2}$$
and so the new series is:
$$\sum_{n=1}^\infty \left(\frac{1}{2n} – \frac{1}{4(n+1)} + \frac{1}{2(n+2)}\right)$$
Not sure how to calculate the sum from here. A solution would be appreciated!
Best Answer
There is an error in your calculation. $B$ should be $-1$, then \begin{equation*} \begin{aligned} \sum_{n=1}^\infty \frac{1}{n^3+3n^2+2n}&=\sum_{n=1}^\infty \left(\frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)}\right)\\ &=\frac{1}{2}\sum_{n=1}^\infty \left[\left(\frac{1}{n} - \frac{1}{n+1}\right) - \left(\frac{1}{n+1} - \frac{1}{n+2}\right)\right]\\ &=\frac{1}{2}\left[\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots\right)-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\cdots\right)\right]\\ &=\frac{1}{2}\left(1-\frac{1}{2}\right)\\ &=\frac{1}{4}. \end{aligned} \end{equation*}