Obliterate me

$\begin{aligned}\begin{aligned}\displaystyle\sum_{k=1}^\infty\frac{k^2-\frac12}{k^4+\frac14}=&\displaystyle\sum_{k=1}^\infty\frac{4k^2-2}{4k^4+1}\\=&\displaystyle\sum_{k=1}^\infty\frac{2(2k^2-1)}{(2k^2-2k+1)(2k^2+2k+1)}\\=&\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\left(\frac{2k-1}{2k^2-2k+1}-\frac{2k+1}{2k^2+2k+1}\right)\\=&\displaystyle\lim_{n\to\infty}\left(1-\cancel{\frac35}+\cancel{\frac35}-\cancel{\frac5{15}}+\cancel{\frac5{15}}-\cancel{\frac7{25}}+\cdots+\cancel{\frac{2n-1}{2n^2-2n+1}}-\frac{2n+1}{2n^2+2n+1}\right)\\=&\displaystyle\lim_{n\to\infty}\left(1-\frac{2n+1}{2n^2+2n+1}\right)\\=&\displaystyle\lim_{n\to\infty}1-\displaystyle\lim_{n\to\infty}\frac{\frac2n+\frac1{n^2}}{2+\frac2n+\frac1{n^2}}\\=&1-0\\=&\boxed{1}\end{aligned}\end{aligned}$