For example, suppose I had the series
\begin{equation} f(x)=\sum_{m=0}^{\infty}\frac{(-1)^m}{m!}x^m. \end{equation}
We know this is the power series expansion for $e^{-x}$, which is square integrable on $[0,\infty]$. Suppose I did not know the closed form of the power series above. Is there any way to determine if $f(x)$ is square integrable (on $[0,\infty]$) from the coefficients of the power series? If square integrability cannot be determined, is it at least possible to know if $f(x) \rightarrow 0$ as $x \rightarrow \infty$?
Another example, suppose I have a power series solution from a linear second order ODE. Following the Frobenius method, suppose my solution to this ODE is of the form
\begin{equation} g(x)= x^r \sum_{m=0}^{\infty}a_mx^m, \end{equation}
with a complicated recurrence relation for the $a_m$ and $r$ is any real number. Assume the closed form of the series is unknown or does not exist, but it converges absolutely for all $x$. Can I determine if $g(x)$ is square integrable on $[0,\infty]$ or if $g(x) \rightarrow 0$ as $x \rightarrow \infty$ from the power series alone?