Let $ \Omega = \{z ∈ 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 $\lim_{n\to \infty }a_n = −c$, where $c$ is the residue of $f$ at $z=1$.
At every point $z_0;|z_0|<1$ ;$f$ has a Taylor series expansion $f(z) =\sum _{n=0}^\infty a_nz_n$; where $a_n=\dfrac{n!}{2\pi i}\int _\gamma \dfrac{f(z)}{(z-z_0)^{n+1}} dz$ where $z_0$ lies in the interior of $\gamma$.
How can I take the limit of $\lim_{n\to \infty }a_n$ from this expression ?Is it at all possible?
Please help.Only hints needed.
Define the function $g(z) = (z-1)f(z)$, which is analytic on $\Omega$, because of the assumptions you have on $f$. Compute $$ g(z) = \sum_{n=0}^{\infty}a_{n}z^{n+1} - \sum_{n=0}^{\infty} a_n z^n = -a_0 + \sum_{n=1}^{\infty}(a_{n-1} - a_n)z^n$$ Since $g(1)$ exists, the serie $\Sigma_n (a_{n-1}-a_n) $ must converge ; thus $(a_n)$ converges and $\lim\limits_{n \to \infty} a_n = -c$