I have the following problem,
$$\sum_{k=1}^\infty\frac{1}{n(n+1)(n+2)}$$
And I try to work as follows:
Hint: Partial Fraction decomposition:
$\begin{aligned} \frac{1}{n(n+1)(n+2)} &= \frac{a}{n}+\frac{b}{(n+1)}+\frac{c}{(n+2)}\\ &= \frac{a(n+1)(n+2)}{n}+\frac{b(n)(n+2)}{(n+1)}+\frac{c n(n+1)}{(n+2)}\\ &= \frac{(a+b+c)n^2+(a3+2b+c)n+2a}{n(n+1)(n+2)} \end{aligned}$
$\begin{alignat}{7} a &&\; + \;&& b &&\; - \;&& c &&\; = \;&& 0 & \\ 3a &&\; + \;&& 2b &&\; + \;&& c &&\; = \;&& 0 & \\ 2a &&\; + \;&& 0b &&\; + \;&& 0c&&\; = \;&& 1 & \end{alignat} $
$$a = \frac{1}{2}; b = -1; c = \frac{1}{2};$$ $$\frac{1}{n(n+1)(n+2)} =\frac{1}{2}(\frac{1}{n}+\frac{1}{(n+2)})-\frac{1}{(n+1)}$$
what follows:
$$\sum_{k=1}^\infty\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\sum_{k=1}^\infty(\frac{1}{n}+\frac{1}{(n+2)})-\sum_{k=1}^\infty\frac{1}{(n+1)}$$
And now, how I can continue?
Note: I found a possible solution in sum of infinite series with telescopes but I can not find a clear answer. Can you help me? Thanks!
As a summary after the answers:
We have
$\frac{1}{n(n+1)(n+2)} =\frac{1}{2}(\frac{1}{n}+\frac{1}{(n+2)})-\frac{1}{(n+1)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\left(\frac{1}{(n+2)}-\frac{1}{(n+1)}\right) $
Consequently: $$\sum_{k=1}^\infty\frac{1}{n(n+1)(n+2)} =\frac{1}{2}\underbrace{\sum_{k=1}^\infty\left(\frac{1}{n}-\frac{1}{(n+1)}\right)}_{\text{telescoping sums}}+\frac{1}{2}\underbrace{\sum_{k=1}^\infty\left(\frac{1}{(n+2)}-\frac{1}{(n+1)}\right)}_{\text{telescoping sums}}= $$
$$\frac{1}{2}\cdot\lim\limits_{N \rightarrow \infty}{(1-\frac{1}{N+1})}+\frac{1}{2}\cdot\lim\limits_{N \rightarrow \infty}{(\frac{1}{N+2}-\frac{1}{2})} =\frac{1}{2} - \frac{1}{4} = $$
$$\sum_{k=1}^\infty\frac{1}{n(n+1)(n+2)} = \frac{1}{4}$$
Thank you all for your answers!
