Tangent series representation

How to prove that for any complex number $z$ which is not equal to $\pi k + \frac{\pi}{2}$ ($k\in\mathbb Z$) : $$ \tan z = \sum_{n=0}^\infty \frac{8z}{(2n+1)^2\pi^2 - 4z^2} $$ Using complex analysis, I started with the contour intergal $$ \oint_{C_N} \frac{\tan \frac{\pi s}{2}}{s^2-z^2}\,\mathrm ds = \sum_{n=-N}^N \frac{-4i}{(2n+1)^2 - z^2} + \frac{2\pi i \tan \frac{\pi z}{2}}{z}$$ where $C_N$ is the circle centered at 0 of radius $N+1/2$ ($N\in\mathbb N$).
The complex number $z$ is chosen to be non zero & non odd integer.

However, I don't know how proceed to show that the LHS goes to $0$ as $N\to \infty$ :(

Thanks in advance for answers.

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You can use the Mittag-Leffler Expansion:

$\ds{\tan\pars{z}}$ has single poles at $\ds{p_{n} = \pars{2n + 1}{\pi \over 2}}$, with residues $\ds{r_{n} = -1}$, where $\ds{n \in \mathbb{Z}}$ .

$$ \bbx{\mbox{Note that}\quad p_{-n} = -p_{n - 1}} $$

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Then, \begin{align} \tan\pars{z} & = \sum_{n = -\infty}^{\infty}\pars{-1}\pars{{1 \over z - p_{n}} + {1 \over p_{n}}} = \sum_{n = 1}^{\infty}\bracks{% \pars{{1 \over p_{n} - z} - {1 \over p_{n}}} + \pars{{1 \over p_{-n} - z} - {1 \over p_{-n}}}} \\[2mm] & + \pars{{1 \over p_{0} - z} - {1 \over p_{0}}} \\[5mm] & = \lim_{N \to \infty}\sum_{n = 1}^{N}\bracks{% \pars{{1 \over p_{n} - z} - {1 \over p_{n}}} + \pars{{1 \over -p_{n - 1} - z} + {1 \over p_{n - 1}}}} + \pars{{1 \over p_{0} - z} - {1 \over p_{0}}} \\[5mm] & = \lim_{N \to \infty}\bracks{\pars{{1 \over p_{0} - z} - {1 \over p_{0}}} + \sum_{n = 1}^{N}\pars{{1 \over p_{n} - z} - {1 \over p_{n}}} + \sum_{n = 1}^{N}\pars{{1 \over -p_{n - 1} - z} + {1 \over p_{n - 1}}}} \\[5mm] & = \lim_{N \to \infty}\bracks{\sum_{n = 0}^{N - 1} \pars{{1 \over p_{n} - z} - {1 \over p_{n}}} + \pars{{1 \over p_{N} - z} - {1 \over p_{N}}} + \sum_{n = 0}^{N - 1}\pars{-\,{1 \over p_{n} - z} + {1 \over p_{n}}}} \\[5mm] & = \sum_{n = 0}^{\infty}\pars{{1 \over p_{n} - z} - {1 \over p_{n} + z}} = \sum_{n = 0}^{\infty}{2z \over p_{n}^{2} - z^{2}} = \sum_{n = 0}^{\infty}{8z \over \pars{2p_{n}}^{2} - 4z^{2}} \\[5mm] & = \bbx{\sum_{n = 0}^{\infty}{8z \over \pars{2n + 1}^{2}\pi^{2} - 4z^{2}}} \end{align}