The mean-value property for holomorphic functions states that if f is holomorphic in the neighbourhood of the closed disc centered at $z0$ of radius $R$, then
$f(z_{0}) = \frac{1}{2\pi}\int_{0}^{2\pi}f(z_{0} + re^{i\theta})\, d\theta$
if $f(z)=\sum_{n=0}^{+\infty}a_nz^n$, We have by dominated convergence:
$\int_{0}^{2\pi}f(z_{0} + re^{i\theta})d\theta = \int_{0}^{2\pi}\sum_{n=0}^{+\infty}a_n(z_{0} + re^{i\theta})^nd\theta=\sum_{n=0}^{+\infty}a_n\int_{0}^{2\pi}(z_{0} + re^{i\theta})^nd\theta =[\frac{(z_0+re^{i\theta})^{n+1}}{ir(n+1)}]_0^{2\pi}=0$
Which means evry holomorphic function is equal to 0, a non sense. I can't find my mistake, any help much appreciated.
HINT:
Your antiderivative is incorrect. Use the binomial theorem to write
$$(z_0+re^{i\theta})^n =\sum_{k=0}^n \binom{n}{k} z_0^{n-k}e^{ik\theta}$$
then exploit the fact that $\int_0^{2\pi}e^{ik\theta}\,d\theta=0$ for $k\ne 0$.