Consider the function $f(z)=1+z+z^2+z^3+\ldots$. This series is absolutely convergent on the disc $|z|<1$ and is equal to $1/(1-z)$ in this region. Now, $1/(1-z)$ is a meromorphic function on $\mathbb{C}$ with a simple pole at $z=1$. Is this sufficient to show that $f$ itself has a meromorphic continuation to the whole complex plane with a simple pole at $z=1$?
Intuitively and "formally" this seems correct. But I am not sure if there are any analytic arguments hidden in the background that I am missing.
Thank you.
The function $$f(z)=1+z+z^2+z^3+\ldots$$ is well defined for $D = \{ z \in \mathbb C | |z|<1 \}$.
The function $$g(z)=1/(1-z)$$ is analytic on on $\mathbb C \setminus \{1\}$.
On the domain $D$, $f(z) = g(z)$ so $g$ satisfies the definition of analytic continuation of $f$ to $\mathbb C \setminus \{1\}$, this continuation is unique (consider $f(z)-g(z)=0$).