Limit of a quotient of factorials 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? 
 A: 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
$$\lim_{n\to\infty}\frac{((2n)!)^k}{(2nk+k-1)!}=0$$
A: 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 $\approx 6.97\times 10^{-414}$
A: $\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}
$
