I try to solve the following problem but have some difficulties:
Let $\Omega=\{z \in \mathbb{C}: |z|<2 \}$ and $\ f$ be a function on $\Omega$ which is holomorphic at every point of $\Omega$ except $z=1$ and at $z=1$ it has a simple pole. Suppose that $$f(z)=\sum_{n=0}^{\infty}a_nz^n, \ |z|<1.$$ Prove that $\lim_{n \rightarrow \infty}a_n=-c,$ where $c$ is the residue of $\ f$ at $z=1$.
First of all, I know that $c=\ Res(\ f(z),1 ) = \frac{1}{2 \pi i}\int_{\gamma}f(z) dz$, where $\gamma$ is any rectifiable curve around $1$ in $\Omega$. Now if we expand $\ f$ in the power series in $|z|<1$ (as given in the problem) then the coefficients $a_n's$ are given by $a_n=\frac{f^{(n)}(0)}{n!}=\frac{1}{2 \pi i}\int_{\Gamma} \frac{f(z)}{z^{n+1}} dz.$ Here $\Gamma$ is any rectifiable curve inside $|z|<1$ . I thought to use the Cauchy Estimates to conclude about $a_n's$. But this idea didn't work (I think)...
But I can't see any hint to capture the problem from here. Any suggestion and hint is appreciated. Thank You.
Let $\varphi(z)=(z-1)f(z)$ for $z\ne 1$ and $\varphi(1)=c$ for $z=1$, then $\varphi$ is holomorphic throughout $\Omega$. Since $\varphi$ has a representation of $(z-1)\displaystyle\sum_{n=0}^{\infty}a_{n}z^{n}=-a_{0}+\sum_{n=1}^{\infty}(a_{n-1}-a_{n})z^{n}$ on $\{|z|<1\}$, this series expansion must hold throughout $\Omega$. Now $\varphi(1)=-a_{0}+\displaystyle\sum_{n=1}^{\infty}(a_{n-1}-a_{n})$ exists, and by telescoping we have $c=\lim_{n\rightarrow\infty}(-a_{n})$.