Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$.. Question : 
Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$ 
What I have done : 
nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2)$ respectively.
Therefore the nth term of given series is : 
$\frac{2r-1}{r(r+1)(r+2)} =\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}$ .....(1)
By using partial fraction : 
and solving for A,B and C we get A = 1/2, B = -1, C =1/2
Putting the values of A,B and C in (1) we get : 
$\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)}$
But by putting $r =1,2,3, \cdots$  I am not getting the answer. Please guide how to solve this problem . Thanks.
 A: The $r$th term is
$$\frac12 \left (\frac1{r}-\frac1{r+1} \right )-\frac12 \left (\frac1{r+1}-\frac1{r+2} \right )$$
so the sum telescopes.  The result is
$$\sum_{r=1}^n \frac{2 r-1}{r (r+1) (r+2)} = \frac12\left (1-\frac1{2} \right )-\frac12 \left (\frac1{n+1}-\frac1{n+2} \right ) = \frac14 - \frac1{2 (n+1) (n+2)}$$
A: HINT:
If the $r$th term
$$T_r=\frac1{2r}-\frac1{r+1}+\frac1{2(r+2)}$$
$$2T_r=\frac1r-\frac2{r+1}+\frac1{(r+2)}=\left(\underbrace{\frac1r-\frac1{r+1}}\right)-\left(\underbrace{\frac1{r+1}-\frac1{r+2}}\right)$$
Recognize the two Telescoping series
A: You can get the result to pop out if you play around with the indices of the infinite series and expand a couple terms. $$\sum_{r=1}^\infty\frac{2r-1}{r(r+1)(r+2)} = \sum_{r=1}^\infty\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)} \\ = \frac{1}{2}\sum_{r=1}^\infty \frac{1}{r}-\sum_{r=1}^\infty\frac{1}{r+1}+\frac{1}{2}\sum_{r=1}^\infty\frac{1}{(r+2)} \\ = \frac{1}{2}\left(1+\frac{1}{2}+\sum_{r=3}^\infty \frac{1}{r}\right)-\left(\frac{1}{2}+\sum_{r=3}^\infty\frac{1}{r}\right)+\frac{1}{2}\sum_{r=3}^\infty\frac{1}{r} \\ = \frac{1}{2}\left(1+\frac{1}{2}\right)-\frac{1}{2}+\sum_{r=3}^\infty \frac{1}{r}\left(\frac{1}{2}-1+\frac{1}{2} \right)$$ The answer should be clear from here.
