Let $f$ be an analytic function in the annulus $ \{ 0<|z|<R \}$
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?
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 $\left\{0<\left|z\right|<R\right\}$ 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 $n$.