How to calculate $\sum_{r=1}^\infty\frac{8r}{4r^4+1}$?

We can write the sum as:

$$\sum _{r=1}^{\infty} \frac{8r}{4r^4+1}$$

$$= \sum _{r=1}^{\infty} \frac{8r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}$$

$$= \sum _{r=1}^{\infty} \frac{2}{2r^2 - 2r + 1} - \frac{2}{2r^2 + 2r + 1}$$

$$= \sum _{r=1}^{\infty} \frac{2}{2r^2 - 2r + 1} - \frac{2}{2(r + 1)^2 - 2(r + 1) + 1}$$

$$=\frac{2}{2\cdot1^2 - 2\cdot1 + 1}$$

$$ = 2$$

Hint

$$4(r)^4+1=(2r^2+1)^2-(2r)^2=(2r^2-2r+1)(2r^2+2r+1)$$

and use fraction partial decomposition.

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?