Let $f_n$ be a uniformly bounded sequence of analytic functions on $\Omega\subset\mathbb C$.
If $f_n(z)\to f(z)$ forall $z\in\Omega$, then by the Montel's theorem I know that the convergence is uniform on compact sets, hence $f$ is an analytic function.
Assume instead only that $$ \frac{1}{n}\sum_{k=1}^nf_k(z) \to f(z) $$ for all $z\in\Omega$. May I say that $f$ is analytic?
Yes, The sequence $T_n (z) =\frac{1}{n} \sum_{j=1}^n f_j (z)$ is also uniformly bounded sequence of analytic functions.