On the evalution of an infinite sum

$$ \begin{align}\newcommand{\Re}{\operatorname{Re}} &\sum_{n=0}^\infty(-1)^n\left[\frac{2n+1/2}{(2n+1/2)^2+1}+\frac{2n+3/2}{(2n+3/2)^2+1}\right]\tag1\\ &=\frac12\sum_{n\in\mathbb{Z}}(-1)^n\left[\frac{2n+1/2}{(2n+1/2)^2+1}+\frac{2n+3/2}{(2n+3/2)^2+1}\right]\tag2\\ &=\frac14\sum_{n\in\mathbb{Z}}(-1)^n\left[\frac{n+\frac14}{\left(n+\frac14\right)^2+\frac14}+\frac{n+\frac34}{\left(n+\frac34\right)^2+\frac14}\right]\tag3\\ &=\frac18\sum_{n\in\mathbb{Z}}(-1)^n\left[\frac1{n+\frac14-\frac i2}+\frac1{n+\frac14+\frac i2}+\frac1{n+\frac34-\frac i2}+\frac1{n+\frac34+\frac i2}\right]\tag4\\ &=\frac18\left[\frac\pi{\sin\left(\pi\!\left(\frac14-\frac i2\right)\right)}+\frac\pi{\sin\left(\pi\!\left(\frac14+\frac i2\right)\right)}+\frac\pi{\sin\left(\pi\!\left(\frac34-\frac i2\right)\right)}+\frac\pi{\sin\left(\pi\!\left(\frac34+\frac i2\right)\right)}\right]\tag5\\ &=\frac{\pi\sqrt2}8\left[ \frac{\cosh\left(\frac\pi2\right)+i\sinh\left(\frac\pi2\right)}{\cosh(\pi)}+ \frac{\cosh\left(\frac\pi2\right)-i\sinh\left(\frac\pi2\right)}{\cosh(\pi)}+ \frac{\cosh\left(\frac\pi2\right)-i\sinh\left(\frac\pi2\right)}{\cosh(\pi)}\right.\\ &\left.\phantom{=\frac{\pi\sqrt2}8}+ \frac{\cosh\left(\frac\pi2\right)+i\sinh\left(\frac\pi2\right)}{\cosh(\pi)}\right]\tag6\\ &=\frac\pi{\sqrt2}\frac{\cosh(\pi/2)}{\cosh(\pi)}\tag7 \end{align} $$ Explanation:
$(2)$: use symmetry
$(3)$: pull factor of $\frac12$ out front
$(4)$: partial fractions
$(5)$: use $(3)$ from this answer
$(6)$: evaluate the sine of a complex number
$(7)$: simplify