Is there a good approximation for this?

Question

What is a good approximation for $\dfrac{k!}{\binom{k^2}{k}}$ as a function of $k$?

Is there a $k_0\in\Bbb N$ such that for all $k\gt k_0$, $$\frac{c}{2^{dk}}\gg\frac{k!}{\binom{k^2}{k}}\gg\frac{c}{2^{dk(\log_2 k)^{1+a}}}$$ with some fixed $c,d>0$ at every $a>0$?

Do we also have $$\frac{c}{2^{dk}}\gg\sum_{i=2}^{k}{\dfrac{(k+1)!}{{i(k+1-i)!\binom{k^2}{i}}}}\gg\frac{c}{2^{dk(\log_2 k)^{1+a}}}$$ with some fixed $c,d>0$ at every $a>0$?

Answer 1

Using almost the same approach as Ross Millikan used $$A=\frac{k!}{\binom{k^2}{k}}=\frac {(k!)^2(k^2-k)!}{(k^2)!}$$ $$\log(A)=2\log(k!)+\log\big((k^2-k)!\big)-\log\big((k^2)!\big)$$ I used Stirling expansion $$\log(n!)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log(n) \right)+\frac{1}{12 n}-\frac{1}{360 n^3}+O\left(\left(\frac{1}{n}\right)^{7/2}\right)$$ Replacing and expanding again as a series for large values of $k$, I arrived to $$\log(A)=-2 k+\left(\frac{1}{2}+\log (2k\pi )\right)-\frac{1}{6 k}-\frac{1}{6 k^2}-\frac{7}{180 k^3}+O\left(\left(\frac{1}{k}\right)^4\right)$$

For $n=10$ the exact result is $\approx 2.096323009\times 10^{-7}$ while the approximation leads to $\approx 2.096324531\times 10^{-7}$.

For $n=20$ the exact result is $\approx 8.725204600\times 10^{-16}$ while the approximation leads to $\approx 8.725205028\times 10^{-16}$.

Edit

As a very first approximation, we should then have $$A=\frac{k!}{\binom{k^2}{k}} \approx 2k \pi\,e^{-2k+\frac 12}$$

Answer 2

First, expand the binomial into factorials. $$\frac{k!}{\binom{k^2}{k}}=\frac {(k!)^2(k^2-k)!}{(k^2)!}$$ Then use Stirling-I will leave incorporating the $\sqrt{2 \pi n}$ to you-it will be small. $$\frac{k!}{\binom{k^2}{k}}=\frac {(k!)^2(k^2-k)!}{(k^2)!}\\ \approx \frac{k^{2k}(k^2-k)^{(k^2-k)}e^{(k^2)}}{(k^2)^{(k^2)}e^{(2k+k^2-k)}}\\=\frac {(k^2-k)^{(k^2-k)}}{k^{2k^2-k}e^k}\\=\frac{(k^2)^{(k^2-k)}(1-\frac 1k)^{(k^2-k)}}{k^{2k^2-k}e^k}\\ =\frac {(1-\frac 1k)^{(k^2-k)}}{k^{k}e^k}$$Can you argue that the numerator is $e^{-k}$ for large $k$?