TITLE: How to find the Laurent expansion for $1/\cos(z)$
QUESTION [5 upvotes]: How to find the Laurent series for $1/\cos(z)$ in terms of $(z-\frac{\pi}{2})$ for all $z$ such that $0<|z-\frac{\pi}{2}|<1$

REPLY [3 votes]: To get a recurrence for the series of the reciprocal, consider
$$
\begin{align}
1
&=\sin(x)\sum_{k=0}^\infty a_kx^{2k-1}\\
&=\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}\sum_{k=0}^\infty a_kx^{2k-1}
\end{align}
$$
and apply Cauchy products to get
$$
\sum_{k=0}^n(-1)^k\frac{a_{n-k}}{(2k+1)!}=\left\{\begin{array}{}
1&\text{if }n=0\\
0&\text{otherwise}
\end{array}\right.
$$
So that we get $a_0=1$ and
$$
a_n=\sum_{k=1}^n(-1)^{k-1}\frac{a_{n-k}}{(2k+1)!}
$$
which gives $\{a_k\}$ to be
$$
\left\{1,\frac16,\frac7{360},\frac{31}{15120},\dots\right\}
$$
Then we get
$$
\begin{align}
\frac1{\cos(x)}
&=-\frac1{\sin(x-\pi/2)}\\
&=-\sum_{k=0}^\infty a_k(x-\pi/2)^{2k-1}\\
&=-\frac1{x-\pi/2}-\frac{x-\pi/2}{6}-\frac{7(x-\pi/2)^3}{360}-\frac{31(x-\pi/2)^5}{15120}-\dots
\end{align}
$$