TITLE: Infinite sum converging to 2
QUESTION [9 upvotes]: How do I compute
$$\sum_{r=1}^{\infty} \frac{8r}{4r^4 +1}$$
Calculating first few terms tells me that the sum converges to 2.
I have also tried squeezing the term.

REPLY [4 votes]: Hint: use that 
$$4r^4+1=(2r^2)^2+1=((2r^2)^2+4r^2+1)-4r^2=(2r^2+1)^2-(2r)^2=(2r^2+2r+1)(2r^2-2r+1)$$
Now, using simple fractions,
$\dfrac{8r}{4r^4+1}=\dfrac{8r}{(2r^2+2r+1)(2r^2-2r+1)}=\dfrac{2}{2r^2-2r+1}-\dfrac{2}{2r^2+2r+1}=2\left(\dfrac{1}{2r^2-2r+1}-\dfrac{1}{2r^2+2r+1}\right)$.
The idea is calculate $\sum\limits_{r\in\Bbb N}\frac{1}{2r^2\pm 2r+1}$ using the fact that $$\frac{1}{2r^2\pm 2r+1}=\int^1_0 x^{2r^2\pm 2r}dx$$ and interchanging the sum with the integral.
An example is here: Finding the infinite sum of a rational function using integrals