Let $f$ be holomorphic in $|z|<2$ except for $z=1$, in which it has a simple pole. Let $f(z)=\sum_{n=0}^{\infty} a_n z^n$ be the Taylor series of $f$ around $z=0$. Prove or find a counterexample:
(A) $a_n$ doesn't necessarily converge.
(B) $a_n$ converges and $\lim_{n\to \infty} a_n = -Res(f,1)$.
(C) $\sum_{n=0}^{\infty} a_n$ converges.
(D) $a_n\neq 0 $ for every $n$ and $\sum_{n=-\infty}^0 \frac{1}{a_{-n}} z^n$ is the Laurent series of $f$ in $1<|z|<2$.
(E) There exists a $z_0$ with $|z_0|>1$ for which $\sum _{n=0}^\infty$ converges.
The only parts I know how to solve are: (C) is wrong since it represents the value of the Taylor series at $z=1$ which we know doesn't converge because the function has a simple pole there.
(B) looks true, but I don't know how to approach it. I guess there is some use of the Residue theorem, not sure how to prove that $\frac{1}{2\pi i }\int \frac{f(z)}{z^{n+1}}dz \to -Res(f,1)$ as $n$ tends to infinity.
In $D$ I thought that in such a case $z=1$ would be an essential singularity, but wasn't sure.
Thank you!