How do you determine that the series solution to the hermite differential equation is not square integrable?

Question

When solving the Schrodinger equation of the harmonic oscillator in one dimension you encounter the hermite differential equation:

\begin{equation} \left[\frac{d^{2}H}{d\xi^{2}}-2\xi\frac{d H}{d \xi }+\left(\lambda-1\right) H\right]e^{-\xi^{2}/2}=0. \end{equation}

We can find a solution to this equation in the form of a series:

\begin{equation} H(\xi)=\sum_{n=0}^{\infty} a_{n}\xi^{n} \ \end{equation}

with the following coefficients: \begin{equation} a_{n+2}=\frac{\left[2n-\left(\lambda-1\right)\right]a_{n}}{(n+2)(n+1)}. \label{coefficient_série} \end{equation} Since we look for a square integrable solution we usually truncate the series by imposing that $\lambda=2n+1$. My question is how do you assess that$\langle\psi|\psi\rangle$ diverges, with $\psi$= $H(\xi)e^{-\frac{\xi^2}{2}}$?

Usually in textbooks it's said that $H(\xi)$ behave assymptotically for large $\xi$ as $e^{\xi^2}$. Does someone as the actual derivation of this behaviour? From my understanding the only thing we can say is that for large value of $n$ we have:

\begin{equation} \frac{a_ {n+2}}{a_{n}} \simeq \frac{2}{n} \end{equation}

\begin{equation} e^{\xi^{2}}=\sum_{n=0}^{\infty}\frac{\xi^{2n}}{n!}=\sum_{n=0}^{\infty}c_{n}\xi^{2n} \end{equation}

\begin{equation} \left|\frac{c_{n+1}}{c_{n}}\right|=\frac{n!}{(n+1)!}=\frac{1}{n+1}. \end{equation}

So we see that both series are convergent with an infinite radius of convergence and that in both series the coefficients decreases in $\frac{1}{n}$. How is it sufficient to say that $H(\xi)$ behave as $e^{\xi^2}$ when $\xi$ goes to infinity?

Answer 1

For large $n$, $$\frac{a_{n+2}}{a_{n}}\approx\frac{2}{n}$$ means $$a_n\approx\frac{C}{(n/2)!}$$ for some constant $C$ and therefore $$H(\xi)\approx C\sum_n \frac{1}{(n/2)!} \xi^n\approx C\sum_l\frac{1}{l!}\xi^{2l}\approx C\exp(\xi^2)$$

Edit: To make this rigorous, it's enough to get a lower bound $H(\xi)\ge\epsilon \exp(\xi^2)$ for some $\epsilon>0$ and large enough $\xi$. That's enough to make the solution non-square-integrable.