# Convergence of Laurent Series.

Let $f(z)$ be analytic for $|z|>r$ and let it be bounded $|f(z)|\leq M, M>0$ wherever it is analytic. Show that the coefficients of the Laurent Series of $f(z)$ are $0$ for $j\geq 1$.

$$\sum_{j=0}^\infty a_j(z-z_0)^j$$
converges when $|z-z_0|<R$. In our case $R$ is infinity so the only way for the sum to converge is if $a_j=0$. Does this make sense?
The function $F(z) = f(1/z)$ is holomorphic for $0 < |z| < 1/r$, where it is uniformly bounded by $M$. So $F$ has a removable singularity at $0$, which gives a power series expansion $$F(z) = \sum_{n=0}^{\infty}b_nz^n,\;\;\; 0 < |z| < 1/r.$$ Therefore, $$f(z) = F(1/z) = \sum_{n=0}^{\infty}b_n z^{-n},\;\;\; |z| > r.$$ Your argument doesn't work because you're not addressing boundedness. A series can convergence everywhere in $\mathbb{C}$.