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## Main Question or Discussion Point

Suppose we have

[tex]f(z) = \sum_{n=-\infty}^\infty c_n z^n; \quad U(z) = \sum_{n=0}^\infty c_n z^n; \quad L(z) = \sum_{n=1}^\infty c_{-n} z^{-n}[/tex]

where [itex]f(z)[/itex] converges* in the interior of some annulus with inner radius [itex]r[/itex] and outer radius [itex]R > r[/itex]. Further suppose [itex]U(z)[/itex] has radius of convergence [itex]R_0[/itex] and [itex]L(1/z)[/itex] has radius of convergence [itex]1/r_0[/itex], so [itex]L(z)[/itex] converges for [itex]|z| > r_0[/itex]. Must it be the case that [itex]r_0 \leq r < R \leq R_0[/itex]? Obviously the sum over all integers converges for [itex]r_0 < |z| < R_0[/itex] (if any such z exist), but for instance what happens if [itex]R_0 < r_0[/itex]? It's conceivable that the upper and lower sums both individually diverge, but that their combined sum still converges. It doesn't seem possible to me for this case to actually occur on a non-empty open set like an annulus... but I dunno.

For instance, where does

[tex]... + \frac{1}{2^3} \frac{1}{z^3} + \frac{1}{2^2} \frac{1}{z^2} + \frac{1}{2} \frac{1}{z} + 1 + z + z^2 + z^3 + ...[/tex]

converge?

This question came about in the context of the uniqueness of Laurent series. It seems that Laurent series are typically assumed to only be defined where their upper and lower sums converge, which sidesteps this issue. Each source I've looked at (Rudin's Real and Complex Analysis; Wikipedia; MathWorld; this book) at best includes this requirement implicitly and at worst ignores it entirely which makes me wonder if it's really necessary.

*I'm interpreting [itex]\sum_{n=-\infty}^\infty c_n z^n[/itex] as [itex]\lim_{N \rightarrow \infty} \sum_{n=-N}^N c_n z^n[/itex].

[tex]f(z) = \sum_{n=-\infty}^\infty c_n z^n; \quad U(z) = \sum_{n=0}^\infty c_n z^n; \quad L(z) = \sum_{n=1}^\infty c_{-n} z^{-n}[/tex]

where [itex]f(z)[/itex] converges* in the interior of some annulus with inner radius [itex]r[/itex] and outer radius [itex]R > r[/itex]. Further suppose [itex]U(z)[/itex] has radius of convergence [itex]R_0[/itex] and [itex]L(1/z)[/itex] has radius of convergence [itex]1/r_0[/itex], so [itex]L(z)[/itex] converges for [itex]|z| > r_0[/itex]. Must it be the case that [itex]r_0 \leq r < R \leq R_0[/itex]? Obviously the sum over all integers converges for [itex]r_0 < |z| < R_0[/itex] (if any such z exist), but for instance what happens if [itex]R_0 < r_0[/itex]? It's conceivable that the upper and lower sums both individually diverge, but that their combined sum still converges. It doesn't seem possible to me for this case to actually occur on a non-empty open set like an annulus... but I dunno.

For instance, where does

[tex]... + \frac{1}{2^3} \frac{1}{z^3} + \frac{1}{2^2} \frac{1}{z^2} + \frac{1}{2} \frac{1}{z} + 1 + z + z^2 + z^3 + ...[/tex]

converge?

This question came about in the context of the uniqueness of Laurent series. It seems that Laurent series are typically assumed to only be defined where their upper and lower sums converge, which sidesteps this issue. Each source I've looked at (Rudin's Real and Complex Analysis; Wikipedia; MathWorld; this book) at best includes this requirement implicitly and at worst ignores it entirely which makes me wonder if it's really necessary.

*I'm interpreting [itex]\sum_{n=-\infty}^\infty c_n z^n[/itex] as [itex]\lim_{N \rightarrow \infty} \sum_{n=-N}^N c_n z^n[/itex].