Sum of a series to be attempted by telescopic method. Ok now this one is pretty tough, my sir said that this couldn't be done by convention...
$$ \sum_{r=1}^n\left(\frac{r^2-\frac12}{r^4+\frac14}\right)$$ 
Ok, so applying the same sense in this question, I cannot go ahead. The denominator and the numerator don't match up to give me something. I believe it is interesting though. Help me.
 A: By partial fraction decomposition we have
$$\frac{r^2-\frac{1}{2}}{r^4+\frac{1}{4}}=\frac{1}{2}\left(\frac{4r-2}{2r^2-2r-1}-\frac{4(r+1)-2}{2(r+1)^2-2(r+1)-1}\right)=\frac{1}{2}(u_r-u_{r+1})$$
so by telescoping we have
$$ \sum_{r=1}^n\left(\frac{r^2-\frac12}{r^4+\frac14}\right)=\frac{1}{2}\sum_{r=1}^n(u_r-u_{r+1})=\frac{1}{2}(u_1-u_{n+1})$$
A: Hint:
$$\begin{align}  r^4+\frac{1}{4}&=\left(r^2+\frac{1}{2}\right)^2-r^2\\&= \left(r^2+\frac{1}{2}-r\right)\left(r^2+\frac{1}{2}+r\right)\\&=\left(r^2+\frac{1}{2}-r\right)\left[(r+1)^2+\frac{1}{2}-(r+1)\right]\end{align}$$
A: In general, any rational function summation with a possibility of telescoping summation trick typically admits a partial fraction decomposition where you can express the decomposition of the $n$th term as a difference of a function $f(n) - f(n+1)$ or at least as $f(n) - f(n+k)$ for some integer $k \geq 1$ (although if $k > 1$ this makes the resulting formula you get from telescoping trick more complicated, but still possible to write down with $k$ cases)
