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Let $f$ be an analytic function in the annulus $ \{ 0<|z|

Is it true that the Laurent series expansion of $f$ about $0$ has a finite number of negative coefficients?

It came to my mind since $0=r=\limsup_{n\rightarrow\ \infty}|a_{n}|^{1/n}$ and therefore $|a_{n}|$ must be $0$ at some point.

Am I correct?

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First a Laurent expansion is defined in an annulus and thus the statement: "Laurent expansion about $0$" makes little sense. If you are refering to the Laulernt expansion at $\left\{0<\left|z\right| then it depends:

If $f$ has a removable singularity at $0$, there won't be any negative powers.

If $f$ has a pole there, there will be finitely many negative powers.

If $f$ has an essential singularity there, there will be infinitely many negative powers.

The coeffecients may very well be negative.

Oh and your proof is incorrect. $\lim\sup\left|a_n\right|^{\frac1n}=0$ doesn't imply $a_n=0$ for large $n$.

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    @GEdgar or even $e^{1/z}$2012-12-24