Summation of series Find $\sum_1^n$ $\frac {2r+1}{r^2(r+1)^2}$ 
Also, find the sum to infinity of the series.
I tried decomposing it into partial fractions of the form $\frac Ar$ + $\frac{B}{r^2}$ + $\frac{C}{(r+1)}$ + $\frac{D}{(r+1)^2}$ but it was getting too complicated and tedious. Is there some trick here that i'm missing?

$$
\frac1{r^2} - \frac1{(r+1)^2} = \frac{(r+1)^2-r^2}{r^2(r+1)^2} = \frac{(r^2+2r+1)-r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2}
$$
The thing that suggested this to me is that $\displaystyle 2r+1 = 2\left(r+\frac12\right)$, and $r+\dfrac12$ is half-way between $r+0$ and $r+1$, the two expressions in the denominator.
Next, you rely on the fact that the sum telescopes, so nearly all of the terms vanish.
