Divergence of Mehler's Hermite polynomial series

Question

According to Mehler's formula, $$ \sum\limits_{n=0}^{\infty}\left(\cfrac{\rho}{2}\right)^n\cfrac{H_n(x)H_n(y)}{n!}=\cfrac{1}{\sqrt{1-\rho^2}}\exp\left(-\cfrac{\rho^2(x^2+y^2)-2\rho x y}{1-\rho^2}\right), $$ when $|\rho|<1$, where $H_n(x)$ are Hermite polynomials. My questions is about what happens when $|\rho|> 1$. Is it possible to show that the series on the left-hand goes to $+\infty$ for all $x$ and $y$? What happens when $|\rho|=1?$.

Answer 1

Let $x, y$ be fixed reals. It is known that asymptotically for $n$ large,

\begin{cases} H_n(x) = \frac{\Gamma(n+1)}{\Gamma(\frac{n}{2}+1)} e^{\frac{x^2}{2}} + k_n \qquad \qquad \;\; \text { if $n$ even } \\ H_n(x) = \frac{\Gamma(n+2)}{\sqrt{2n+1}\Gamma(\frac{n}{2}+\frac{3}{2})} e^{\frac{x^2}{2}} + k_n \qquad \text { if $n$ odd } \end{cases}

where $| k_n | \leq \frac{M}{\sqrt{2n+1}}$, for some constant $M > 0$. This asymptotic approximation is uniform on $[x, y]$ (Olver et al. NIST Handbook of Mathematical functions, 2010 edition, eqs. 8.15.24-8.15.27 p. 453).

As $\frac{\Gamma(n+1)}{\Gamma(\frac{n}{2}+1)} \sim \sqrt{2} \left(\frac{n}{e}\right)^{n+1}\left(\frac{n}{2e}\right)^{-\frac{n}{2}-1}\sim 2^{\frac{n+3}{2}}\left(\frac{n}{e}\right)^{\frac{n}{2}}$ we therefore have $H_n(x) \sim 2^{\frac{n+3}{2}}\left(\frac{n}{e}\right)^{\frac{n}{2}} e^\frac{x^2}{2} $ for large values of $n$ even and a similar expression for $H_n(y)$, so that for $n$ even :

$$\frac {H_n(x)H_n(y)}{n!} \sim \frac{2^{n+3}}{\sqrt{2\pi n}} e^\frac{x^2+y^2}{2}$$

whence, still for $n$ even:

$$ \frac {H_n(x)H_n(y)}{n!} \left(\frac{\rho}{2}\right)^n \sim \frac{2^{\frac{5}{2}}e^{\frac{x^2+y^2}{2}}}{\sqrt{\pi}} \frac{\rho^n}{\sqrt{n}}$$

By the ratio test and the divergence of the series $\;\sum_{n=0}^\infty \frac{1}{\sqrt{n}}$, this sub-series of even terms therefore absolutely converges if and only if $|\rho| < 1$. It is not necessary to investigate odd terms: as these are necessarily positive for all $n$ large enough, divergence of the sub-series suffices to infer divergence of the whole series.

This is what had to be shown.