Let consider a Laurent series $\displaystyle{ \sum_{k\in\mathbb Z}a_kz^k }$ with complex coefficients and converging inside the annulus $A=\{\ z\in\mathbb C\ |\ r<|z|<R\ \}$, with $r<1<R$. We denote by $f(z)$ the sum of the series on $A$. To start, I don't have any assumption on the convergence on the boundary of the annulus.
I am interested in the sum $\displaystyle{S_n=\frac{1}{2^n}\sum_{n=0}^{2^n-1}f\left(e^\frac{2k\pi}{2^n}i\right)}$. It is easy to check that $2\pi S_n$ is the Riemann sum of the continuous function $F(t)=f(e^{it})$ for the equipartite subivision in $2^n$ parts of $[0,2\pi]$, hence $\displaystyle{ \lim_{n\to\infty} S_n=\frac{1}{2\pi}\int_{0}^{2\pi}F(t)\ dt}$.
A parametrization shows that $\displaystyle{ \frac{1}{2\pi}\int_{0}^{2\pi}F(t)\ dt}$ is equal to $\displaystyle{ \frac{1}{2i\pi}\int_{C} \frac{f(z)}{z}\ dz }$ (where $C$ is the unit circle in $\mathbb C$), which in turn is equal to $a_0$ by the Residue Formula.
Therefore we proved
Proposition: $$\displaystyle{\lim_{n\to\infty}\frac{1}{2^n}\sum_{n=0}^{2^n-1}f\left(e^\frac{2k\pi}{2^n}i\right)=a_0}$$
What happens to the limit of $S_n$ when $r=R=1$? Of course, if the series diverges at any point point of $C$ (e.g $\displaystyle{ \sum_{k\in\mathbb Z}z^k }$), $f$ is not even defined.
To precise my question, I am looking for answers in 2 cases:
- The case where the series converges on all $C$ (e.g $\displaystyle{ \sum_{k\in\mathbb Z, k\neq 0} \frac{z^k}{k^2} }$)? Will the above proposition still be true? If yes, how to prove it? Can I still expect $f$ to be continuous (so that my proof still holds)?
- The case where the series converges for all roots of unity (or at least all the roots of $z^{2^n}=1$).