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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).

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up vote 3 down vote accepted

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.

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You're absolutely right, thank you. But what if I need the series expansion of $g(x)$ only where the sum converges in the first place? Even better: what if I only need the first term of the series? Isn't there a way to derive that from the functions $f_n (x)$? – Joe Jun 27 '11 at 8:28
You can expand g(x) around a point in the region where the sum converges. So in your example, if you take it around $x=1$, $f_n(x) = e^{-n} - 2 n e^{-n} (x-1) + n (2n - 1) e^{-n}(x-1)^2 + \ldots$, and then you can sum each term over $n$ to get the expansion of $g(x)$ in powers of $x-1$. – Robert Israel Jun 27 '11 at 13:43
Then I guess that if I want to expand $g(x)$ around a point the sum doesn't converge in, even if I want the expansion to hold only in a region where the sum does converge, I'm out of luck. Oh well, thanks anyway! – Joe Jun 28 '11 at 6:38
I have another thought: the problem is in expanding $g(x)$ around a point which is on the boundary of the region in which the sum converges. However, in the example I gave the closed-form expression for $g(x)$ is defined also in regions where the sum doesn't converge (it only has poles at $x^2=2\pi n$ for integer $n$). Thus we can have a series expansion around $x=0$ because it is just an isolated pole. So what I'm asking now is: is there way to learn something from the functions $f_n(x)$ about the analytical continuation of $g(x)$ to regions where the sum doesn't converge? – Joe Jul 1 '11 at 10:37

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