How to solve $\sum _{k=1}^{n}{{4k^2-2}\over{4k^4+1}} $

Question

$$\sum _{k=1}^{n}{{4k^2-2}\over{4k^4+1}} $$

Now this appears to be a telescopic summation and I have reduced it to $\sum _{k=1}^{n}{{4k^2-2}\over(2k^2-2k+1)(2k^2+2k+1)}$ . But after this, I am unable to think of any other manipulation to make a telescopic form. Is this approach even correct or is it a dead end... Please help.

Answer 1

Hint. We have the identity $$\frac{4x^2-2}{4x^4+1}=\frac{2x-1}{2x^2-2x+1}-\frac{2x+1}{2x^2+2x+1}\,.$$

Answer 2

We have $$ \begin{aligned} &\sum_{1\le k\le n} \frac{4k^2-2}{(2k^2-2k+1)(2k^2+2k+1)} \\ &\qquad= 2\sum_{1\le k\le n} \left( \frac{k^2}{2k^2-2k+1} - \frac{(k+1)^2}{2k^2+2k+1} \right) \\ &\qquad= 2\sum_{1\le k\le n}(f(k)-f(k+1)) \\ &\qquad=2(f(1)-f(n+1)) \ , \end{aligned} $$ where $f(k)=k^2/(2k^2-2k+1)$.