My question is about a series representation of $\sec(z)$ where $z$ is a complex number, however I don't think complex analysis should have bearing on the mistake being made, so to help make it more familiar, we'll just put things in terms of x.
So I start with the series representation of $\csc(x)$:
$$\sum_{n=-\infty}^\infty \frac{(-1)^n}{x-n\pi}$$
and we want to split the sum up into it's positive and negative parts:
$$\sum_{n=-\infty}^{-1} \frac{(-1)^n}{x-n\pi} + \frac{1}{x} + \sum_{n=1}^\infty \frac{(-1)^n}{x-n\pi}$$
and $\sum_{n=-\infty}^{-1} \frac{(-1)^n}{x-n\pi} = \sum_{n=1}^\infty \frac{(-1)^n}{x-(-n)\pi}$
$$\frac{1}{x}+\sum_{n=1}^\infty (-1)^n\Big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\Big)$$ and we finally end up with:
$$\tag{2}\frac{1}{x}+2x\sum_{n=1}^\infty \frac{(-1)^n}{x^2-(n\pi)^2}$$
So here's where I am getting stuck. I want to change (2) to a series representation of $\csc(x)$ by utilizing $\sec(x) = \csc(x+\pi/2)$. Supposedly by making this change, we arrive to the series representation of $\sec(x)$ as: $$\tag{3}\sum_{n=0}^\infty (-1)^n\frac{ (2n+1)\pi}{(n+\frac{1}{2})^2\pi^2-x^2}$$
Whenever I plug in $x+\pi/2$ into (2), I do not arrive at (3). Some things I noticed from looking at (3), I can see that the denominator is of the form $a^2-b^2$ so I can see immediately tell that it is composed of $\frac{1}{\big(x+\frac{\pi}{2}+n\pi\big)\big(-x+\frac{\pi}{2}+n\pi\big)}$, however if I plug in $x+\pi/2$ into (2) and factor it, I would get a denominator composed of $\frac{1}{(x+\frac{\pi}{2}+n\pi)(x+\frac{\pi}{2}-n\pi)}$.
NOTE: we can also represent $\csc(x) = -\sec(x-\pi/2)$. I was thinking maybe there was a way to apply both this and $\csc(x+\pi/2)$? I don't know, I'm just throwing out ideas.
EDIT: Looks like THIS PROBLEM might be helpful. It has a different starting series, but perhaps the series representation I'm using instead to represent the Mittag-Leffler expansion is not a good one to work to the final form?
Also, Something I noticed that I think can help get rid of the $\frac{1}{x}$ term as well as start the series at $n= 0$ is by incorporating it into the series and approximating it to 0. To show that, let's say $x=1$, then the series would be equal to $$1 + 2 \sum_{n=1}^\infty (-1)^n \frac{1}{-(n\pi)^2}$$ and technically, we can set the sum to start at $n=0$ by simply subtracting the $1/x$ term to get $$(2 \sum_{n=1}^\infty (-1)^n \frac{1}{-(n\pi)^2})-1$$ and we can technically also incorporate this into the sum itself by adding an equal fraction of $1/x$ onto every single term by splitting it up into the number of terms+1. So for example: $$a_0+\sum_{n=1}^4 n = a_0 + (1+2+3+4) \\ = \sum_{n=0}^4 n a_0/(4+1) = (a_0/5+0)+(a_0/5+1)+(a_0/5+2)+(a_0/5+3)+(a_0/5+4) = (a_0/5+a_0/5+a_0/5+a_0/5+a_0/5)+(0+1+2+3+4) = a_0 + (1+2+3+4)$$. So applying this to (2), we would get $$2x\lim_{N\rightarrow \infty}\sum_{n=0}^N \frac{1}{x^2-(n\pi)^2} + \frac{1}{x}\frac{1}{N+1} \approx 2x\sum_{n=0}^\infty \frac{1}{x^2-(n\pi)^2}$$ or at least I think that's a viable way to approximately remove the term at $n=0$ while also setting the series to $n=0$. The point is that the $n=0$ term split into smaller equal parts added onto every other term becomes insignificantly small as the number of number of terms in the series goes to infinite.
