Q:

Laurent series for an infinite sum of functions

I have an infinite sum of analytic functions that is guaranteed to converge for every $x$, except for $x=0$:
\begin{equation}
g(x) = \sum_{n=1}^\infty f_n (x)
\end{equation}
I want to expand the function $g(x)$ in a Laurent series around $x=0$.
For example if $f_n(x)=e^{-nx^2}$, then:
\begin{equation}
g(x)=\frac{1}{e^{x^2}-1} \approx \frac{1}{x^2}-\frac{1}{2}+\frac{x^2}{12} ...
\end{equation}
The problem is that in general I don't have a closed-form expression for $g(x)$.
However, I think I should still be able to obtain a closed-form expression for the coefficients of the Laurent series.  Taylor-expanding $f_n(x)$ is not good enough, since the terms of the Taylor series do not necessarily converge in $n$ (in the above example they don't). 

A:

When dealing with analytic functions, you really need to consider complex $x$, not just real $x$.  In your example the series $\sum_{n=1}^\infty e^{-n  x^2}$ converges only when $\text{Re } x^2 > 0$.  In any example where the series converges [EDIT: uniformly on compact subsets ] in a punctured disk $\{z: 0 < |z| < \epsilon\}$ you can indeed take the Laurent series term-by-term.   If convergence is only in some sectors,  I doubt that much can be said in general, especially if you want closed-form solutions. 