Summation of general series

By computing the residues of $\frac{x^3}{x^4+4}$ in $x=\pm 1\pm i$ we have: $$\frac{x^3}{x^4+4}=\frac{1}{2}\left(\frac{x-1}{(x-1)^2+1}+\frac{x+1}{(x+1)^2+1}\right)\tag{1}$$ hence: $$\begin{eqnarray*}\sum_{n=0}^{k}\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}&=&\frac{1}{2}\sum_{n=0}^{k}(-1)^n\left(\frac{2n}{4n^2+1}+\frac{2(n+1)}{4(n+1)^2+1}\right)\\&=&\frac{(-1)^k\cdot(k+1)}{4(k+1)^2+1}.\end{eqnarray*}$$