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.
Best Answer
We can write the sum as:
$$\sum _{r=1}^{\infty} \frac{8r}{4r^4+1}$$
$$= \sum _{r=1}^{\infty} \frac{8r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}$$
$$= \sum _{r=1}^{\infty} \frac{2}{2r^2 - 2r + 1} - \frac{2}{2r^2 + 2r + 1}$$
$$= \sum _{r=1}^{\infty} \frac{2}{2r^2 - 2r + 1} - \frac{2}{2(r + 1)^2 - 2(r + 1) + 1}$$
$$=\frac{2}{2\cdot1^2 - 2\cdot1 + 1}$$
$$ = 2$$