On a quiz I saw the equation: $$a_{n}=\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{n!}$$ and I was asked to use a graph of it in order to guess if it does converges or diverges as $n \to \infty$, and if it converge to "guess" $\lim_{ n \to \infty } a_{n}$.
The graph did diverge and I put that down as the answer, but the wording was interesting to me, so I tried to find the limit to see if it did. I came up with $$\lim_{ n \to \infty } a_{n} =\lim_{ n \to \infty }\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{1\cdot2\cdot3\cdot4\cdot5\cdot\ldots\cdot n}= \lim_{ n \to \infty }\frac{1}{2\cdot4\cdot6\dots n}$$ The numerator is every odd number multiplied together, and the denominator is every natural number multiplied together, so if they cancel out, it should just be 1/(product of every even natural number)
Giving this to Wolfram Alpha gave me $$\lim_{n\to\infty}\frac{\prod_{x=1}^{n}(2x-1)}{n!}=\infty$$ I'm aware that rearranging the addends in infinite sums can change it, is that also true for infinite products? Is what I did correct or is there something I'm missing
If you want to have a better idea of how fast $a_n$ grows, I'd do the following. First, note that \begin{aligned} a_n &= \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{1\cdot 2\cdot 3 \cdots n} \\ &= \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{1\cdot 2\cdot 3 \cdots n} \cdot \frac{2 \cdot 4 \cdot 6 \cdots (2n-2)}{2 \cdot 4 \cdot 6 \cdots (2n-2)} \\ &= \frac{(2n-1)!}{n!} \cdot \frac{1}{(2\cdot 1) \cdot (2\cdot 2) \cdot (2\cdot 3) \cdots (2(n-1))} \\ &= \frac{(2n-1)!}{n!} \cdot \frac{1}{2^{n-1}(1\cdot 2 \cdot 3 \cdots (n-1))} \\ &= \frac{(2n-1)!}{2^{n-1}n!(n-1)!}. \end{aligned}
Now, we can use Stirling's large-$n$ approximation, that is $$ \ln(n!) \approx n\ln(n) -n + \frac{1}{2}\ln(n)+\frac{1}{2}\log(2\pi) $$ to show that, after some algebra, \begin{aligned} \ln(a_n) &= \ln\bigl((2n-1)!\bigr) - \ln(n!) - \ln\bigl((n-1)!\bigr) - (n-1)\ln(2) \\ &\approx n\ln(2) - \frac{1}{2}\ln(n)-\frac{1}{2}\ln(\pi) \end{aligned} or, equivalently, $$ a_n \approx \frac{2^n}{\sqrt{n\pi}}. $$