Finding $\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{2r^2-1}{4r^4+1}$
TRY $4r^4+1=(2r^2-2r+1)(2r^2+2r+1)$ So $\displaystyle\sum^{n}_{r=1}\frac{2r^2-1}{(2r^2-2r+1)(2r^2-2r+1)}$ but how i convert in Telescopic sum because in numeratoror it has $(2r^2-2)$ instead of $4r$. Thanks
On
This is too long for a comment.
In the initial post, there was a "small" typo (namely $(2r^2-1)$ in the title but $(2r^2-2)$ in the body of the question).
This made me considering the more general problem of $$A_k=\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{2r^2+k}{4r^4+1}$$
What I did was to consider $a,b,c,d$ to be the roots of $4r^4+1=0$ and use partial fraction decomposition leading to $$\frac{2r^2+k}{4r^4+1}=\frac \alpha {r-a}+\frac \beta {r-b}+\frac \gamma {r-c}+\frac \delta {r-d}$$ and, using generalized harmonic numbers $$\sum^{n}_{r=1}\frac 1{r-a}=H_{n-a}-H_{-a}$$ Now, considering the asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ recombining, simplifying (a lot of tedious work - but I enjoyed), we finally end with the nice $$A_k=-\frac{k}{2}+\frac{\pi}{4} (k+1) \tanh \left(\frac{\pi }{2}\right)$$ which shows the specificity of $k=-1$.
Hint: $\;\; \displaystyle\frac{2r^2-1}{(2r^2-2r+1)(2r^2+2r+1)} = \frac{1}{2}\left(\frac{2r-1}{2r^2-2r+1} - \frac{2r+1}{2r^2+2r+1}\right) \,$.