Finding the lowest value for $a$ at which $\lim_{n\to\infty}\frac{(2n)!}{(n!)^a}$ is convergent 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
 A: 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.
A: 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 $0$.

