How to calculate $\sum_{r=1}^\infty\frac{8r}{4r^4+1}$? 
Calculate the following sum: $$\frac{8(1)}{4(1)^4+1} + \frac{8(2)}{4(2)^4+1} +\cdots+ \frac{8(r)}{4(r)^4+1} +\cdots+ \text{up to infinity}$$

MY TRY:- I took $4$ common from the denominator. and used $a^2+b^2=(a+b)^2-2ab$. It gave me two brackets, whose subtraction was written in numerator. so I did the same thing as we do in the method of partial fraction, and started putting $1,2,3$ and so on. my answer came didn't match with the right answer.

Let the general term be $\frac {8r}{4r^4+1}=\frac {2r}{r^4+(\frac {1}{2})^2}=\frac {2r}{(r^2+\frac {1}{2})^2-r^2} =\frac {2r}{(r^2+\frac {1}{2}-r)(r^2+\frac {1}{2}+r)} $  now $2r=r^2+\frac {1}{2}+r-(r^2+\frac {1}{2}-r) $ so iur series is a telescoping one and is equal to $S=\sum _0^{\infty} \frac {1}{r^2+\frac {1}{2}-r}-\frac {1}{r^2+\frac {1}{2}+r}=2$
