Maclaurin series for $e^z /\cos z$. I want to find the Maclaurin series for the function
$$f(z)=\frac{e^z}{\cos z}.$$
Right away I can tell that the radius of convergence will be $\pi/2$, since it's the distance to the nearest singularity (not sure if this explanation is rigorous enough, but I can't think of anything else). Calculating the coefficients as n'th derivatives at $0$ strikes me as fruitless here, so I'm guessing I have to resort to tricks. Of course I could just divide the series formally but that doesn't get me an explicit formula for the coefficient $a_n$.
I tried rewriting the cosine as an exponential, but that too produces nothing of interest. I'm looking for a hint to get me started.
 A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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 \newcommand{\ds}[1]{\displaystyle{#1}}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
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$$
\expo{z} = \sum_{k = 0}^{\infty}{z^{k} \over k!}
\quad
\left\vert%
\begin{array}{rcl}
\qquad\sec\pars{z} & = & \sum_{k = 0}^{\infty}{\verts{E_{2k}} \over \pars{2k}!}z^{2k}
\\
\verts{z} & < & {\pi \over 2}
\\[2mm]
&&E_{\nu}\ \mbox{is an}\ {\it\mbox{Euler Number}}.
\end{array}\right.
$$
\begin{align}
&\vphantom{\Huge A^{A}}
\\
{\expo{z} \over \cos\pars{z}}
&=
\expo{z}\sec\pars{z}
=
\sum_{k = 0}^{\infty}\sum_{k' = 0}^{\infty}{z^{k} \over k!}\,
{\verts{E_{2k'}} \over \pars{2k'}!}z^{2k'}
=
\sum_{k = 0}^{\infty}\sum_{k' = 0}^{\infty}{z^{k} \over k!}\,
{\verts{E_{2k'}} \over \pars{2k'}!}
\sum_{n = 0}^{\infty}z^{n - k}\,\delta_{n,k + 2k'}
\\[3mm]&=
\sum_{n = 0}^{\infty}z^{n}
\sum_{k' = 0 \atop {\vphantom{\LARGE A}n - 2k' \geq 0}}^{\infty}
{1 \over \pars{n - 2k'}!}\,
{\verts{E_{2k'}} \over \pars{2k'}!}
\end{align}

$${\large%
{\expo{z} \over \cos\pars{z}}
=
\sum_{n = 0}^{\infty}A_{n}\,z^{n}\,,
\qquad\qquad
A_{n}
\equiv
\sum_{k = 0}^{\floor{n/2} \atop }
{\verts{E_{2k}} \over \pars{2k}!\pars{n - 2k}!}}
$$

