If the complex functions $f_0,\,f_1,\,f_2,\,\ldots$ are holomorphic in the disc $\vert z-z_0\vert < r$ and thus

(1)

(2)

(3)

(4)

Proof. Apparently, the series (2) converges uniformly also in every closed sub-disc of the open disc $|z-z_0| < r$ . Therefore the theorem 2 in the entry ``theorems on complex function series'' says that the sum $F(z)$ is holomorphic in $|z-z_0| < r$ and $$F^{(\nu)}(z) = f_0^{(\nu)}(z_0)+f_1^{(\nu)}(z_0)+f_2^{(\nu)}(z_0)+\ldots \quad (\nu = 0,\,1,\,2,\,\ldots).$$ Theorem 3 in the same entry thus guarantees that $F(z)$ has the Taylor expansion of the form (4) wherein $$A_\nu = \frac{1}{\nu!}F^{(\nu)}(z_0) \quad (\nu = 0,\,1,\,2,\,\ldots).$$ According to theorem 2 in the same entry the series (2) may be differentiated termwise, $$A_\nu = \frac{1}{\nu!}\sum_{n=0}^\infty f_n^{(\nu)}(z_0) = \sum_{n=0}^\infty \frac{1}{\nu!}f_n^{(\nu)}(z_0) = \sum_{n=0}^\infty a_{n\nu}$$ Q.E.D.

Note. In Weierstrass double series theorem it's a question of changing the summing order: