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I would like to prove the following conjecture:

$$\lim_{n \to \infty} \frac{t^{2n+2}e^{-\frac{1}{2}t^2}}{(2n+1)!!}=0$$ for all $t \in \mathbb R$. Naturally, this can only be done by showing that $(2n+1)!! \gt t^{2n+2}$ as $n \to \infty$. Unfortunately, I have no idea how to approach this problem. The presence of the independent variable $t$ complicates the matter; while it may be easy to show that the above equation is true for a specific value of $t$, my problem involves a rigorous proof of the conjecture for all real values of $t$.

[Also, $n \in \mathbb N$.]

Note that this conjecture is assumed to be true in an article entitled "Solution for the Indefinite Integral of the Standard Normal Probability Density Function" by Joram Soch. The proof for this conjecture is not included the article.

RadiantDank
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  • Well since $$(2n+1)!!=\frac{(2n+1)!}{2^n n!}$$

    And $(2n+1)!>2^n \cdot n! \cdot t^{2n+2}$ seems highly unlikely (Desmos suggests as much), I'd suspect you need the $e^{\frac{-t^2}{2}}$ term for the proof.

     – Rhys Hughes Feb 16 '19 at 22:45

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Hint: note that $t^{2n+2} = a^{n+1}$, where $a = t^2$. So it suffices to show that $\lim\limits_{n\to \infty}\frac{a^n}{(2n+1)!!} = 0$ for any $a \geq 0$. Since $(2n+1)!! \geq n!$ for all $n \in \mathbb{N}$ (both are a product of $n$ terms but the left-hand side has each term greater than the corresponding term on the right-hand side), we have $0\leq \frac{a^n}{(2n+1)!!} \leq \frac{a^n}{n!}$. Hence it suffices to show that $$\lim\limits_{n\to \infty}\frac{a^n}{n!} = 0\quad \forall a \geq 0.$$

Have a go at proving this last one. The basic idea is that there are an equal number of terms being multiplied in the numerator and denominator. But given any $a\geq 0$, eventually $n$ will overtake say $2a$ and we will be always be multiplying by terms bigger than $2a$ in the denominator. Some proofs if you're stuck can be found at Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.

Minus One-Twelfth
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