Evaluating Telescopic Sum $ \sum\frac{n}{1+n^2+n^4} $

Question

How to evaluate following $$ \sum_{n=1}^{\infty}\frac{n}{1+n^2+n^4}$$ I posted my way as an answer, Is there another Interesting approach to evaluate this sum of series?

Answer 1

I did it this way

$x$^th Term of this sum can be written as $$T_x=\frac{x}{1+x^2+x^4}=\left(\frac {1}{2\cdot(x^2-x+1)}\right)-\left(\frac{1}{2\cdot(x^2+x+1)}\right)$$ Similarly $(x+1)$^th Term can be written as $$T_{x+1}=\frac{x+1}{1+(x+1)^2+(x+1)^4}=\left( \frac{1}{2\cdot(x^2+x+1)}\right)-\left(\frac{1}{2\cdot(x^2+3x+3)}\right)$$ So, $$\begin{align} \sum_{n=1}^{\infty}\frac{n}{1+n^2+n^4}&=\frac{1}{3}+\frac{2}{21}+\frac{3}{91}+\cdots+\frac{1}{\infty}\\ &= \left(\frac{1}{2}-\frac{1}{6}\right)+ \left(\frac{1}{6}-\frac{1}{14}\right)+ \left(\frac{1}{14}-\frac{1}{26}\right)+ \cdots \left(\frac{1}{\infty}-\frac{1}{\infty}\right)\\ &=\frac{1}{2}+ \left(-\frac{1}{6}+\frac{1}{6}\right)+ \left(-\frac{1}{14}+\frac{1}{14}\right)+ \cdots \left(\frac{1}{\infty}-\frac{1}{\infty}\right)\\ &=\frac{1}{2}+ \left(0\right)+ \left(0\right)+ \cdots \left(0\right)\\ &=\frac{1}{2} \end{align}$$

Hence, $$\sum_{n=1}^{\infty}\frac{n}{1+n^2+n^4}=\frac{1}{2}$$