I'm working on this problem:
Suppose $f$ is analytic on the disk $|z| < 2$ except at $z = 1$, where $f$ has a simple pole. If $$f(z) = \sum_{n=0}^{\infty}a_nz^n$$ is the Taylor series expansion for $f$ around $z = 0$, prove that $\lim\limits_{n\to \infty} a_n$ exists and equals the residue of $f$ at $z = 1$.
Here's what I tried:
Let $P(z) = \displaystyle \frac{A}{z-1}$ be the principal part of $f$ near $z = 1$. Then $$g(z) := f(z) - P(z) = f(z) + \displaystyle \frac{A}{1-z}$$ is analytic in $|z| < 2$ when $z \ne 1$ since both $f$ and $P$ are. It's also analytic at $|z| = 1$ since the Laurent series there now has no principal part.
So $f - P$ is in particular analytic at $z = 0$ and so has a Taylor series there $$g(z) = \sum_{n=0}^{\infty}(a_n + A)z^n$$ Since $g$ is analytic on $|z| < 2$ the Taylor series for $g$ converges on $|z| = 1$. Therefore $\lim\limits_{n\to \infty} (a_n + A) = 0$ and so $\lim\limits_{n\to \infty} a_n = -A$.
So my answer seems to have a different sign than the question suggests it should have, but I don't see the error in my answer.
Thanks for any help.
Your derivation seems correct. The question is incorrectly stated though. Take one of the simplest possible examples, namely $$ f(z) = \frac{1}{1-z} = \sum_{n=0}^\infty z^n $$ Here $\lim_{n\to\infty} a_n = 1$, but the residue of $f$ at $z=1$ is $-1$, not $1$.