<s>[INST] 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.
 [/INST] We can use partial fraction decomposition to write this as \begin{align}\sum_{n=1}^\infty \frac{8n}{4n^4+1}&=\sum_{n=1}^\infty\left(\frac{2}{2 n^2 - 2 n + 1} - \frac{2}{2 n^2 + 2 n + 1}\right)\end{align}
If we start writing out terms of this sequence, we get \begin{array}{ccccccccccc}(n=1)&2&-&\frac{2}{5}\\
(n=2)&&+&\frac 25&-&\frac 2{13}\\
(n=3)&&&&+&\frac 2{13}&-&\frac 2{25}\\
&&&&&&+&\cdots\\
(n=\infty)&&&&&&&+&\frac{2}{2\infty^2-2\infty+1}&-&\frac{2}{2\infty^2+2\infty+1}\end{array}
We therefore have a telescoping sequence, so the sum is:
$$\sum_{n=0}^\infty \frac{8n}{4n^4+1}=2-\frac{2}{2\infty^2+2\infty+1}$$
We can see that, $$\lim_{n\to\infty}\dfrac{2}{2n^2+2n+1}=0$$
Therefore, $$\sum_{n=1}^\infty\frac{8n}{4n^4+1} = 2$$
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