Given the Laurent series $\sum\limits_{k=-\infty}^\infty a_k^{(l)} z^k = f(z)|_{r_l<|z|<R_l}$ of a meromorphic function $f$ on $\mathbb C$ with convergence region $r_l< |z|< R_l$, one can use analytical continuation to obtain the values in the regions $r_m<|z|<R_m$ where either $r_m=R_l$ or $R_m=r_l$, i.e. switch between series converging in neighbouring rings the boundaries of which are touching isolated singularities. But how can the new coefficients $a_k^{(m)}$ be obtained from the old ones?
As an example to what I mean, take the geometric series $$1+q+q^2+... = \sum_{k=0}^\infty q^k = \frac1{1-q}\Bigg|_{|q|<1}$$ which conveges for $|q|<1$, and its "counterpart" for $|q|>1$, $$\frac1{1-q}\Big|_{|q|>1} = \frac1q\frac{1}{\tfrac1q-1}\Bigg|_{\big|\tfrac1q\big|<1} = -\frac1q\sum_{k=0}^\infty \left(\frac1q\right)^k = -\sum_{k=-\infty}^1 q^k$$
So the question for that example would be, how to get from the $r_1=0\le|q|<1=R_1$ coefficients $$a_k^{(1)} = \begin{cases}1 & k\ge0 \\ 0 & k<0\end{cases}$$ to the $r_2=R_1=1<|q|<\infty=R_2$ ones $$a_k^{(2)} = \begin{cases}0 & k > 1 \\ -1 & k\le 1\end{cases}$$ without the trick I used, i.e. only from the coefficients (and convergence radii)?