Limit $n \rightarrow \infty$ of a fraction of successive Hermit polynomials

Question

I was trying to apply the ratio test to

$$ \sum_{n=0}^{\infty} H_n(x) \frac{t^n}{n!} $$

, where $H_n(x)$ is a Hermite Polynomial. Assuming that both $t$ and $x$ are finite, we start with

$$ \lim_{n \rightarrow \infty} \left| \frac{ H_{n+1}(x) t^{n+1} / (n+1)! }{ H_{n}(x) t^{n} / n! } \right| = \lim_{n \rightarrow \infty} \left| \frac{H_{n+1}(x)}{H_{n}(x)} \frac{t}{(n+1)} \right| $$

I know we have the recurrence relations:

$$ H_{n+1}(x) = 2xH_{n}(x) - 2nH_{n-1}(x) \\ H_{n}'(x) = 2nH_{n-1}(x) $$

but I don't see how these help in evaluating the limit.

Question: Evaluate the following limit:

$$ \lim_{n \rightarrow \infty} \left| \frac{H_{n+1}(x)}{H_{n}(x)} \frac{t}{(n+1)} \right| $$

Answer 1

Using the following representation $$H_{n}(x)=\frac{n!}{2\pi{i}}\oint_{|z|=1}\frac{e^{2xz-z^{2}}}{z^{n+1}}$$ You have $$\lim_{n\rightarrow\infty}|\frac{H_{n+1}}{H_{n}(n+1)}|=\lim_{n\rightarrow\infty}|\frac{\frac{(n+1)!}{2\pi{i}}\oint_{|z|=1}\frac{e^{2xz-z^{2}}}{z^{n+2}}}{(n+1)\frac{n!}{2\pi{i}}\oint_{|z|=1}\frac{e^{2xz-z^{2}}}{z^{n+1}}}|=$$ $$=\lim_{n\rightarrow\infty}|\frac{\oint_{|z|=1}\frac{e^{2xz-z^{2}}}{z^{n+2}}}{\oint_{|z|=1}\frac{e^{2xz-z^{2}}}{z^{n+1}}}|=0$$