You can obtain the desired result using absolute convergence and some observations about the annulus of convergence of Laurent series.
If $\{x_n\}_{n \in \mathbb{Z}}$, then let $R_+(\{x_n\}) = ( \limsup_{n \to \infty} \sqrt[n]{|x_n|} )^{-1}$, and $R_-(\{x_n\}) = \limsup_{n \to \infty} \sqrt[n]{|x_{-n}|}$.
Two relevant results (all summations are over $\mathbb{Z}$):
(i) If $\sum_m \sum_n |a_{m,n}| < \infty$, then the summation can be rearranged. In particular, $\sum_m \sum_n a_{m,n} = \sum_k \sum_l a_{l,k-l}$. (Since $\phi(m,n) = (m,m+n)$ is a bijection of $\mathbb{Z}^2$ to $\mathbb{Z}^2$.)
(ii) If $\sum_{n} |x_n| < \infty$, then $R_+(\{x_n\}) \ge 1$. Similarly, $R_-(\{x_n\}) \le 1$.
Let $f(z) = \sum_n f_n (z-z_0)^n$, $g(z) = \sum_n g_n (z-z_0)^n$. Let $R_+ = \min(R_+(\{f_n\}),R_+(\{g_n\}))$, $R_- = \max(R_-(\{f_n\}),R_-(\{g_n\}))$.
Choose $R_-<r < R_+$. Then $\sum_m |f_m| r^m$ and
$\sum_n |g_n| r^n$ are absolutely convergent sequences, and so $\sum_m \sum_n |f_m||g_n| r^{m+n} < \infty$. From (i) we have $\sum_m \sum_n |f_m||g_n| r^{m+n} = \sum_k (\sum_l |f_l| |g_{k-l}|) r^k < \infty$.
If we let $c_k = \sum_l f_l g_{k-l}$, this gives $\sum_k |c_k| r^k < \infty$, and so
$R_+(\{c_kr^k\}) = \frac{1}{r} R_+(\{c_k \}) \ge 1$, or, in other words, $R_+(\{c_k \}) \ge r$. Since $r<R_+$ was arbitrary, it follows that $R_+(\{c_k \}) \ge R_+$. The same line of argument gives $R_-(\{c_k \}) \le R_-$.
Hence we have $c(z) = f(z)g(z) = \sum_n c_n (z-z_0)^k$ on $R_- < |z-z_0| < R_+$, where $c_k = \sum_l f_l g_{k-l}$.
Best Answer
As this is a second order pole, you only need consider $\sum_{n=-2}^{\infty} a_{n}z^{n}$. Now, check the limits of $[z^3\cdot f(z)]''/2!$ and $[z^2\cdot f(z)]'/1!$ as $z \rightarrow 0$ and we get $1/2$ and $0$ respectively. Then, evaluate $\lim\; 1/(1-\cos(z)) - 2/z^2 \rightarrow \frac{(1/4!)}{(1/2!)} = 1/6$ as $z \rightarrow 0$. This is as many terms as I went. But there is no noticeable pattern as far as I can see at a first glance.