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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:

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

1 Answers 1

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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 $$\lim_{n\rightarrow\infty}|\frac{H_{n+1}}{H_{n}(n+1)}|=\lim_{n\rightarrow\infty}|\frac{\frac{(n+1)!}{2\pi{i}}\oint_{|z|=1}\frac{e^{2xz-z^{2}}}{z^{n+2}}}{(n+1)\frac{n!}{2\pi{i}}\oint_{|z|=1}\frac{e^{2xz-z^{2}}}{z^{n+1}}}|=$$ $$=\lim_{n\rightarrow\infty}|\frac{\oint_{|z|=1}\frac{e^{2xz-z^{2}}}{z^{n+2}}}{\oint_{|z|=1}\frac{e^{2xz-z^{2}}}{z^{n+1}}}|=0$$

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    Thank you! The limit at the end is zero because $z^{n+2}$ goes to infinity faster than $z^{n+1}$, right? If this is correct, we do not even need to evaluate the integral. And, therefore, do not need to set $|z|=1$? – 2017-02-20
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    Yes! The choice $|z|=1$ is arbitrary, you actually need any contour enclosing the pole at the origin. And why you were wondering about this problem? I mean this is a well-known series, and it converges to $\exp(2xt-t^{2})$... – 2017-02-20