Let $f$ be meromorphic in $\{z\in\mathbb{C}:|z|\le2\}$ with a simple pole at $z=1$ and no other poles. Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ be the expansion of $f$ valid for $\{z\in\mathbb{C}:|z|<1\}$. Then, can we can we conclude that $\lim_\limits{n\to\infty}a_n=-\text{Res}_{z=1}f(z)$?
I am blown by this problem. I can only think of the definition of residue as $\text{Res}_{z=a}=R$, where $f(z)-\frac{R}{z-a}$ has a derivative. I also think the Laurent series e xpansion has some role to play. Any hints. Thanks beforehand.
We need the further assumption that $f$ has no other pole in the closed unit disk for the conclusion. If $f$ has another pole there, we usually don't have $\lim\limits_{n\to\infty} a_n = -\operatorname{Res}(f;1)$.
If that is the case, then $g \colon z \mapsto f(z) - \frac{R}{z-1}$ is holomorphic on a disk with radius $\rho > 1$. Then its MacLaurin series
$$\sum_{n = 0}^{\infty} b_n z^n$$
converges at $z = 1$, whence $\lim\limits_{n\to\infty} b_n = 0$. But
$$g(z) = f(z) + \frac{R}{1-z} = \sum_{n = 0}^{\infty} a_n z^n + R\sum_{n = 0}^{\infty} z^n = \sum_{n = 0}^{\infty} (a_n + R)z^n$$
for $\lvert z\rvert < 1$, and so $b_n = a_n + R$ for all $n$, whence
$$\lim_{n\to \infty} a_n = \lim_{n\to \infty} (b_n - R) = \Bigl(\lim_{n\to\infty} b_n\Bigr) - R = -R.$$