How to find asymptotic expansion of $\frac{1}{\cos(z)}$ 
Find the asymptotic expantion of coefficients of the exponential generating function $f(z)=\frac{1}{\cos(z)}$ using all of its poles.

My work:
I know that:
$$
\frac{1}{\cos \left( z \right)}=\sum_{n\ge 0}{E_{2n}}\frac{z^{2n}}{\left( 2n \right) !}
$$
To find the poles we solve $\cos(z)=0$ where z is a complex number. Getting, $\chi_k=\pi/2+\pi k$
where $k \in Z$.
Now wrie $\frac{1}{\cos(z)} \sim  \frac{1}{z-\chi_k}= \frac{-1}{\chi_k} \frac{1}{1-z/\chi_k}$.(I'm not sure if this what I am supposed to do).
Then
$$
E_{2n}=-\left( 2n \right) !\sum_{k\in Z}{\frac{1}{\chi _k}}\frac{1}{\chi _{k}^{n}} \\ 
E_{2n}=-\left( 2n \right) !\left( \frac{1}{\left( \frac{\pi}{2} \right) ^{n+1}}+\sum_{k\ge 1}{\left( \frac{1}{\chi _{k}^{n+1}}+\frac{1}{\chi _{k_-}^{n+1}} \right)} \right) 
$$
Then I can look at the internal sum in the last equation and try to simplify it.
Edited:
$f(z)=\frac{1}{cos(z)} \sim \frac{Res(f(z),\chi_k)}{z-\chi_k}$ where $\chi_k=\frac{\pi}{2}+2\pi k$ the
poles of $f(z)$ and  $Res(f(z),\chi_k)= lim_{z\to \chi_k} (z-\chi_k) f(z)$.
After calculations we get that:
$\chi_0= \pi /2$ with Res= -1, $\chi_{-1} = -\pi / 2$ with Res = 1 , $\chi_1=3\pi /2$ with Res = 1, $\chi_{-2}= -3\pi /2$ with Res = -1 etc.
So, $f(z)=\frac{1}{cos(z)} \sim \frac{Res(f(z),\chi_k)}{z-\chi_k} \sim \frac{-1}{z-1/2 \pi}  +\frac{1}{z+1/2 \pi} +\frac{1}{z-3/2 \pi} + \frac{-1}{z+3/2 \pi}+....$
Then if we expand these geometric coulomns and look at their coefficients we get:
$\frac{E_{2n}}{n!} \sim (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} - (2/ 3\pi)^{n+1} +(2/ 3\pi)^{n+1} - (2/ 5\pi)^{n+1} -(-2/ 5\pi)^{n+1}-(2/ 7\pi)^{n+1} +(-2/7\pi)^{n+1}...)$
Thus, $E_{2n} \sim (2n)! (  (2/ \pi)^{n+1} + (-2/ \pi)^{n+1}  - (2/ 5\pi)^{n+1} -(-2/ 5\pi)^{n+1}-(2/ 7\pi)^{n+1} +(-2/7\pi)^{n+1}...)$
Thus $E_{2n} \sim (2n)! (  (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} +O(2/ 5\pi)^{n})$
What do you think.
Any guide will be so appreciated.

You found that
$$
\frac{1}{{\cos z}} = \sum\limits_{k = 0}^\infty  {\left( {\frac{{( - 1)^k }}{{z + \left( {k + \frac{1}{2}} \right)\pi }} - \frac{{( - 1)^k }}{{z - \left( {k + \frac{1}{2}} \right)\pi }}} \right)}  = \sum\limits_{k = 0}^\infty  {( - 1)^k \frac{{(2k + 1)\pi }}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2  - z^2 }}} .
$$
This expansion is actually convergent for all $z \neq (k+1/2)\pi$, $k \in \mathbb{Z}$. If $|z|<\frac{\pi}{2}$, then, by the geometric series,
$$
\frac{1}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2  - z^2 }} = \frac{1}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2 }}\frac{1}{{1 - \left( {z/\left( {k + \frac{1}{2}} \right)\pi } \right)^2 }} = \sum\limits_{n = 0}^\infty  {\frac{{z^{2n} }}{{\left( {k + \frac{1}{2}} \right)^{2n + 2} \pi ^{2n + 2} }}} .
$$
Thus
\begin{align*}
\frac{1}{{\cos z}} & = \sum\limits_{k = 0}^\infty  {( - 1)^k (2k + 1)\pi \sum\limits_{n = 0}^\infty  {\frac{{z^{2n} }}{{\left( {k + \frac{1}{2}} \right)^{2n + 2} \pi ^{2n + 2} }}} } 
\\ &
 = \sum\limits_{n = 0}^\infty  {( - 1)^n \left[ {( - 1)^n \left( {\frac{2}{\pi }} \right)^{2n + 1} 2(2n)!\sum\limits_{k = 0}^\infty  {\frac{{( - 1)^k }}{{(2k + 1)^{2n + 1} }}} } \right]\frac{{z^{2n} }}{{(2n)!}}} ,
\end{align*}
provided $|z|<\frac{\pi}{2}$. Consequently,
$$
E_{2n} = ( - 1)^n \left( {\frac{2}{\pi }} \right)^{2n + 1} 2(2n)!\sum\limits_{k = 0}^\infty  {\frac{{( - 1)^k }}{{(2k + 1)^{2n + 1} }}}
$$
for all $n\geq 0$.
