Limit of the general term of a power series in a pole Let $\Omega = D(0,2)/\{\frac{1}{2}\}$ , where $D(0,2)$ is a disc, $f$ holomorphic in $\Omega$.
$\frac{1}{2}$ is a simple pole for $f$ with residue $1$, calculate $$ lim_{n \to \infty} \frac{f^{n}(0)}{2^{n}n!}$$
Observations : if $f(z) = \sum_{n}a_{n}z^{n}$ in $D(0,\frac{1}{2})$ then $a_{n}=\frac{f^{n}(0)}{n!}$ so $ lim_{n \to \infty} \frac{f^{n}(0)}{2^{n}n!} =  lim_{n \to \infty} a_{n}(\frac{1}{2})^{n}$ . But $\frac{1}{2}$ is a pole so the series is not convergent. How can we proceed?  
 A: We know the principal part of $f$ in $\frac12$, hence we can subtract that to obtain
$$g(z) = f(z) - \frac{1}{z-\frac12},$$
which by removing the removable singularity in $\frac12$ is a holomorphic function in $D(0,2)$,
$$g(z) = \sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!}z^n$$
for $\lvert z\rvert < 2$, in particular the series for $g\left(\frac12\right)$ converges, giving
$$\lim_{n\to \infty} \frac{g^{(n)}(0)}{2^n n!} = 0.$$
It remains to consider the principal part of $f$ in $\frac12$,
$$\begin{align}
\frac{1}{z-\frac12} &= -2\frac{1}{1-2z}\\
&= -2 \sum_{n=0}^\infty 2^nz^n.
\end{align}$$
We thus have $f^{(n)}(0) = g^{(n)}(0) -2^{n+1}n!$ and
$$\lim_{n\to\infty} \frac{f^{(n)}(0)}{2^n n!} = \lim_{n\to\infty} \frac{g^{(n)}(0)}{2^n n!} - 2 = -2.$$
A: On $\Omega$, $\exists $ a holomorphic function $g(z)$ such that $f(z) = \dfrac{g(z)}{(z-1/2)}$ where $g(1/2) \neq 0$. Then $g(z)$ can be represented by a power series converging at least in  $D(0,2)$ and given by $g(z) = \sum_{n=0}^{\infty} \dfrac{g^{(n)}(0)}{n!} z^n$.
Since $g(z) = (z-1/2) f(z)$ we get the following functional relation (for $ n\geq 1$) $g^{(n)}(z) = f^{(n)}(z) (z-1/2) + n f^{(n-1)}(z) \implies g^{(n)}(0) = (-1/2)f^{(n)}(0)  + n f^{(n-1)}(0)$.
From the given residue condition $\lim_{z \rightarrow 1/2} g(z) = 1$. Therefore
$1 = g(0) + \sum_{n=1}^{\infty} [n f^{(n-1)}(0)-\frac{1}{2}f^{(n)}(0)] \frac{1}{n!2^n}$.
You can verify that this leads to a telescoping series and hence we can write,
$ 1 = g(0) + \frac{1}{2} f(0) + \lim_{n \rightarrow \infty} (-1/2) \dfrac{f^{(n)}(0)}{n!2^n}$.
Since $g(0) = -(1/2)f(0)$,
$ \implies \lim_{n \rightarrow \infty} \dfrac{f^{(n)}(0)}{n!2^n} = - 2$.
