An algebra problem by Razing Thunder

The sum $S$ can be rewritten as:

$\begin{aligned} S & = \sum_{n=1}^\infty \frac n{1 + n^2 + n^4} \\ & = \sum_{n=1}^\infty \frac n{(n^2-n+1)(n^2 + n + 1)} \\ & = \sum_{n=1}^\infty \frac 12 \left(\frac 1{n^2-n+1} - \frac 1{n^2+n+1}\right) \\ & = \frac 12 \sum_{n=1}^\infty \left(\frac 1{(n-1)^2 + (n-1)+1} - \frac 1{n^2+n+1} \right) \\ & = \frac 12 \left( \sum_\red{n=0}^\infty \frac 1{n^2 + n+1} - \sum_\blue{n=1}^\infty \frac 1{n^2+n+1} \right) \\ & = \frac 12 \times \frac 1{0^2+0+1} = \frac 12 \end{aligned}$

Therefore $14S = 14 \times \dfrac 12 = \boxed 7$ .

Don't click here if you wanna know the solution , please ignore