# Prove that $\lim_{n\to \infty }a_n = −c$

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?

• You know that $$-c = \lim_{z\to 1} (1-z)f(z),$$ don't you? Jan 19 '17 at 11:39
• @DanielFischer; Since $z=1$ is a simple pole then $f(z)=\dfrac{\phi(z)}{z-1}$ where $\phi(1)\neq 0$ and $\phi$ is analytic at $z=1$ Jan 19 '17 at 13:37
• $c=\text {Res} (f,1)=\lim_{z\to 1} (z-1)f(z)$ Jan 19 '17 at 13:40
• How can I bring $\lim_ {n\to \infty }a_n$ into picture? Jan 19 '17 at 13:41
• $c = \phi(1)$, use the power series expansion of $\phi$, like in the answer below. Jan 19 '17 at 13:42

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 $\lim\limits_{n \to \infty} a_n = -c$
• I get $g(1)=c\implies c=-a_0+\sum_{n=1}^\infty (a_{n-1}-a_n)$ Jan 19 '17 at 13:48
• how do you get $\lim a_n=c$ Jan 19 '17 at 13:50
• @BenStokes What is $-a_0 + \sum_{n = 1}^N (a_{n-1}-a_n)$? Jan 19 '17 at 13:54