The Hermite Gaussians defined by \begin{equation} \psi_n(x)=e^{-x^2/2}H_n(x) \end{equation} where $H_n$ are the Hermite polynomials, span the space of real $L^2$ functions, and one can show that the Gaussian function $e^{-x^2}$ has a Taylor series which converges for all $x$ via the ratio test \begin{equation} r=\lim_{n\to\infty}\bigg\vert\frac{c_{n-1}}{c_n}\bigg\vert=\lim_{n\to\infty}\frac{n}{2x^2}=\infty \end{equation} where $c_n$ is the $n$th term in the Taylor series expansion of $e^{-x^2}$ about $x=0$. Does that mean that given any function $f(x)$ in $L^2$ I can first write it as a sum of Hermite Gaussians, and then write those Hermite Gaussians as infinite sums to get a power series representation of $f(x)$ which converges for all $x$?
If this is correct, then how come the $L^2$ function \begin{equation} g(x)=\frac 1{1+x^2} \end{equation} has a Taylor series about $x=0$ which only converges between $x=-1$ and $x=1$? By writing $g(x)$ as a sum over Hermite Gaussians and then writing those as Taylor series shouldn't I be able to obtain a power series for $g(x)$ which converges for all $x$?
Thanks in advance!