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.

Hint. The sum is telescopic. Note that $4r^4+1=(2r^2-2r+1)(2r^2+2r+1)$, and
$$\frac{8r}{4r^4+1}=\frac{2}{2r(r-1)+1}-\frac{2}{2(r+1)r+1}.$$
Hence
$$\sum_{r=1}^n\frac{8r}{4r^4+1}=\sum_{r=1}^n\left(\frac{2}{2r(r-1)+1}-\frac{2}{2(r+1)r+1}\right)=\frac{2}{2\cdot 1(1-1)+1}-\frac{2}{2(n+1)n+1}.$$
Can you find the sum of the series now?
