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So I've noticed that for larger values of $a$ the function below forms a bell curve shape whereas for lower values it diverges, I would like to find an approximation for the lowest value at which this is true: $$f_a(x)=\frac{(2x)!}{(x!)^a}$$

I have thought about using stirling's approximation: $$n!\approx\sqrt{2\pi n}\left(\frac ne\right)^n$$ which would give me: $$\frac{(2x)!}{(x!)^a}=\frac{\sqrt{4\pi x}\left(\frac{2x}{e}\right)^x}{\left(2\pi x\right)^{a/2}\left(\frac xe\right)^{ax}}=\sqrt{2}\left(2\pi x\right)^{1-a/2}\left(\frac xe\right)^{x(1-a)}2^x$$ $$=2^{x+(3-a)/2}\pi^{1-a/2}x^{x(1-a)+1-a/2}e^{-x(1-a)}$$ if we break this down we are really looking for the convergence of: $$x^{x(1-a)+1-a/2}e^{-x(1-a)}$$ Then my thoughts were that the $x^x$ term seems the most important term, so the answer would just be $a\ge1$ but I feel like this is not correct. Does anyone have any thoughts? Thanks

Henry Lee
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2 Answers2

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When $a=2$ the quotient is $\binom{2n}{n}$ so it goes to infinity. Let $a = 2+\alpha$ with $\alpha>0$

The $$\begin{align} \frac{(2n)!}{(n!)^a}&=\frac1{(n!)^\alpha}\binom{2n}{n}\\ &\sim\frac1{(n!)^\alpha}\frac{(2n)^{2n}e^{-2n}\sqrt{4\pi n}}{n^{2n}e^{-2n}2\pi n}\\ &=\frac1{(n!)^\alpha}\frac{4^n}{\sqrt{\pi n}}\\ \end{align}$$ which goes to $0$ as $n\to\infty$ To see this, take the logarithm and recall that $\log n! \sim n\log n$.

So there is a largest $a$, namely $2$, such that the sequence does not converge, but no smallest $a$ such that it does.

saulspatz
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  • when you say "no smallest $a$ such that it does" if we know the largest $a$ for when it diverges then surely $a>2$ suffices for convergence – Henry Lee Nov 28 '20 at 20:10
  • @HenryLee Yes, that's what I proved. But there is no smallest number for which it converges. If you claim that $2+\varepsilon$ is the smallest, that's wrong, because it also converges for $2+\varepsilon/2$ – saulspatz Nov 28 '20 at 20:12
  • Ah yes I get what you are saying, there is no smallest number after 2 its just any number greater than 2 – Henry Lee Nov 28 '20 at 20:49
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The value you're looking for is $a = 2$.

Indeed, one has, according to Stirling's formula:

$$ \frac{(2n)!}{(n!)^a} \quad \underset{n \to +\infty}{\Large{\sim}} \quad \frac{ \sqrt{2 \pi 2 n} (2n)^{2n} e^{-2n} }{ \left( \sqrt{2 \pi n} (n)^{n} e^{-n} \right)^a } = \sqrt{2} \sqrt{2 \pi}^{1-a} n^{ (2-a)n + \frac{1-a}{2} } e^{(a-2)n} 2^{2n} $$

which you can rewrite as $C \frac{n^{f(n)}}{K^n}$ with $ C := \sqrt{2} \sqrt{2 \pi}^{1-a}$, $K = 4e^{a-2} $ and $f(n) = (2-a)n + \frac{1-a}{2}$.

Then:

  • if $a < 2$, one has $\frac{(2n)!}{(n!)^a} \underset{n \to +\infty}{\longrightarrow} +\infty$.
  • if $a = 2$, the limit is also $+\infty$ because of the term $4^n$
  • if $a > 2$, the limit is $0$.
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