<s>[INST] Infinite sum converging to 2 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. 
 [/INST] Partial fraction expansion gives us$$\begin{align*} & \sum\limits_{r=1}^n\frac {8r}{4r^4+1}=\sum\limits_{r=1}^n\frac 2{2r^2-2r+1}-\sum\limits_{r=1}^n\frac 2{2r^2+2r+1}\\ & =\left[2+\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2-2n+1}\right]-\left[\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2+2n+1}\right]\end{align*}$$Notice how all but the last fraction in the second sum cancels out with the fractions in the first sum. Continuing on indefinitely until the end gives us$$\sum\limits_{r=1}^n\frac {8r}{4r^4+1}=2-\frac 2{2n^2+2n+1}$$As $n\to\infty$, the fraction tends to zero, so your sum equals$$\sum\limits_{r\geq1}\frac {8r}{4r^4+1}=2$$
EDIT: To find the partial fraction decomposition, we first factor the denominator as a product of two quadratics. This can be done by adding and subtracting $4r^2$ so the quartic factors as$$4r^4+1=(2r^2+2r+1)(2r^2-2r+1)$$Now, the decomposition is set up as$$\frac {8r}{(2r^2+2r+1)(2r^2-2r+1)}=\frac {Ar+B}{2r^2-2r+1}+\frac {Cr+D}{2r^2+2r+1}$$
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