Limit of a quotient of factorials

Question

I want to find the limit of $$\lim_{n\to\infty}\frac{((2n)!)^k}{(2nk+k-1)!}$$ for an arbitrary integer $k$. I have done several simulations for $k=5$ or $k=3$, the limit is zero in these cases.

I have tried to proof this, but the $k-th$ power gives me a headache. Is there any way to show this rigorously?

Answer 1

For $k=1$ we have

$$\lim_{n\to\infty}\frac{((2n)!)^k}{(2nk+k-1)!}=\lim_{n\to\infty}\frac{(2n)!}{(2n)!}=1$$

and for $k\ge 2$ by ratio test we have

$$\frac{((2n+2)!)^k}{((2n+2)k+k-1)!}\frac{(2nk+k-1)!}{((2n)!)^k}=\frac{(2n+2)^k(2n+1)^k}{(2nk+3k-1)\ldots(2nk+k)}<$$

$$<\frac{(2n+2)^{2k}}{(2nk+k)^{2k}}\to k^{-2k}<1$$

therefore

$$\lim_{n\to\infty}\frac{((2n)!)^k}{(2nk+k-1)!}=0$$

Answer 2

$\lim_{n\to\infty}\frac{((2n)!)^k}{(2nk+k-1)!} $

Let's start with a simpler, larger quotient using Stirling in the form $n! \approx \sqrt{cn}(n/e)^n$.

$\begin{array}\\ \dfrac{((2n)!)^k}{(2nk)!} &\approx \dfrac{(\sqrt{2cn}(2n/e)^{2n})^k}{\sqrt{2cnk}(2nk/e)^{2nk}}\\ &= \dfrac{(2cn)^{k/2}(2n/e)^{2nk}}{\sqrt{2cnk}(2nk/e)^{2nk}}\\ &= \dfrac{(2cn)^{k/2}(2n)^{2nk}/e^{2nk}}{\sqrt{2cnk}(2nk)^{2nk}/e^{2nk}}\\ &= \dfrac{(2cn)^{k/2}}{\sqrt{2cnk}(k)^{2nk}}\\ &= \dfrac1{\sqrt{2cnk}}\left(\dfrac{(2cn)^{1/2}}{k^{2n}}\right)^k\\ &= \dfrac1{\sqrt{2cnk}}\left(\dfrac{(2cn)^{1/(2n)}}{k^{2}}\right)^{kn}\\ &= \dfrac1{\sqrt{2cnk}}\left(\dfrac{e^{\ln(2cn)/(2n)}}{k^{2}}\right)^{kn}\\ &< \dfrac1{\sqrt{2cnk}}\left(\dfrac{1+\ln(2cn)/(2n)}{k^{2}}\right)^{kn} \qquad\text{since }e^x \le 1+2x\text{ for }0 \le x \le 1\\ &\to 0 \qquad\text{for } k \gt 1\\ \end{array} $

Answer 3

Considering $$A=\frac{((2n)!)^k}{(2nk+k-1)!}\implies \log(A)=k \log((2n)!)-\log((2nk+k-1)!)$$ Now, use Stirling approximation $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({p}\right)\right)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$ Apply it and continue with Taylor expansions to get $$\log(A)=-2 n (k \log (k))-\frac{1}{2} \left(\log \left({k}\right)+(k-1) \left(2 \log (k)-\log \left(\frac{\pi }{n}\right)\right)\right)+O\left(\frac{1}{n}\right)$$ So, basically $$A \sim k^{-2 k n}$$

Edit

Following your question in comments, the next level of approximation would be $$A=\pi ^{\frac{k-1}{2}} k^{-2 k n} \left(\frac{1}{k^2 n}\right)^{k/2} \sqrt{k n}+O\left(\frac{1}{n}\right)$$

Uisng $k=10$ and $n=20$, the first approximation gives $10^{-400}$ while the new one gives $\approx 7.63\times 10^{-414}$ for an exact value $\approx 6.97\times 10^{-414}$