Evaluating $\sum_{k=1}^n\dfrac{k^2-\frac12}{k^4+\frac14}$ [duplicate]

Solution 1:

By partial fraction decomposition we have $$\frac{r^2-\frac{1}{2}}{r^4+\frac{1}{4}}=\frac{1}{2}\left(\frac{4r-2}{2r^2-2r-1}-\frac{4(r+1)-2}{2(r+1)^2-2(r+1)-1}\right)=\frac{1}{2}(u_r-u_{r+1})$$ so by telescoping we have $$ \sum_{r=1}^n\left(\frac{r^2-\frac12}{r^4+\frac14}\right)=\frac{1}{2}\sum_{r=1}^n(u_r-u_{r+1})=\frac{1}{2}(u_1-u_{n+1})$$

Solution 2:

Hint:

$$\begin{align} r^4+\frac{1}{4}&=\left(r^2+\frac{1}{2}\right)^2-r^2\\&= \left(r^2+\frac{1}{2}-r\right)\left(r^2+\frac{1}{2}+r\right)\\&=\left(r^2+\frac{1}{2}-r\right)\left[(r+1)^2+\frac{1}{2}-(r+1)\right]\end{align}$$