The generating function of Hermite's polynomials is given by $G(x,t)=e^{2xt-t^2}$ for $x, t \in \mathbf{R}$. It is known that $\displaystyle G(x,t)=\sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}$ for $x, t \in \mathbf{R}$, where $H_n(x)=(-1)^n e^{x^2} \dfrac{d^n}{dx^n}e^{-x^2}$ for $x \in \mathbf{R}$ -is the $n$-th Hermite's polynomial. By properties of power series, for fixed $x\in \mathbf{R}$ we have that $\frac{\partial G}{\partial t}= \sum_{n=1}^\infty \frac{\partial}{\partial t}\left[H_n(x) \frac{t^n}{n!}\right]=\sum_{n=1}^\infty H_n(x) \frac{t^{n-1}}{(n-1)!}\mbox{ for }t \in \mathbf{R}.$
I have seen in some books that also
\frac{\partial G}{\partial x}= \sum_{n=0}^\infty \frac{\partial}{\partial x}\left[H_n(x) \frac{t^n}{n!}\right]= \sum_{n=0}^\infty H_n'(x) \frac{t^n}{n!} \mbox{ for }x \in \mathbf{R}.
I don't understand, what is the reason that we can differentiate this series with respect to $x$ (maybe series $\sum_{n=0}^\infty \frac{\partial}{\partial x} (...)$ is uniformly convergent or locally uniformly convergent, but I don't see why).
Thanks.