$$ \pi \sec{\left(\frac{\pi x}{2} \right)} = 4 \sum_{k=0}^{\infty} (-1)^k\frac{2k+1}{(2k+1)^2-x^2} . $$
Preferably avoiding complex analysis or zig-zag numbers. This is my missing link in understanding the 1/(4k+1) series. I can get the similar series for tan (by differentiating log(sin) but have become really stuck with this trying similar approaches and looking for trig identities.
Form the sum in the followings:
$S= 4\sum\limits_{k=0}^{\infty} (-1)^k\frac{2k+1}{(2k+1)^2-x^2}=2 \sum\limits_{k=0}^{\infty} \frac{(-1)^k}{2k+1-x}+2\sum\limits_{k=0}^{\infty}\frac{(-1)^k}{2k+1+x}$
Let $k=2m$ if $k$ even and $k=2m+1$ if $k$ odd so we get:
$S=2 \sum\limits_{m=0}^{\infty} \frac{1}{4m+1-x}-2\sum\limits_{k=0}^{\infty}\frac{1}{4m+3-x}+\sum\limits_{m=0}^{\infty} \frac{1}{4m+1+x}-2\sum\limits_{k=0}^{\infty}\frac{1}{4m+3+x}$
Introduce the digamma function: $\psi(1+z)=-\gamma+\sum\limits_{k=1}^\infty\big(\frac{1}{k}-\frac{1}{z+k}\big)$
We have the followings:
$S=\frac{1}{2}\big(-\psi(\frac{x+1}{4})+\psi(\frac{x+3}{4})-\psi(\frac{-x+1}{4})+\psi(\frac{-x+3}{4})\big)$
Using the reflection formula of digamma function:
$\psi(1-z)-\psi(z)=\pi \cot(\pi z)$
We get:
$S=\frac{\pi}{2}\big(\cot(\frac{\pi}{4}(x+1))+\cot(\frac{\pi}{4}(1-x))\big)$
After simplifing the expression using trigonometric laws:
$S=\frac{\pi}{2}\frac{2}{\cos(\frac{x\pi}{2})}=\pi \sec(x\frac{\pi}{2})$
The statement is proved.
