limit to infinite of a factorial number

Question

I have this problem and I solved it like this, but muy teacer told me that it's wrong so I dont know who to do it.

Prove that $\lim _{n \to \infty}\frac{(n!)^22^{2n}}{(2n)!\sqrt{n}}$. I use the Stirling's approximation $$\frac{(n!)^22^{2n}}{(2n)!\sqrt{n}}\approx \frac{((\sqrt{2\pi n})n^ne^{-n})^22^{2n}}{(\sqrt{4\pi n})(2n)^{2n}e^{-2n}\sqrt{n}}=\frac{n^{2n}e^{-2n}2\pi n(2^{2n})}{2^{2n}n^{2n}e^{-2n}2\pi^{0.5}n}=\frac{\pi}{\pi^{0.5}}=\sqrt{\pi}$$ $$\lim _{n \to \infty}\frac{(n!)^22^{2n}}{(2n)!\sqrt{n}}\approx \lim _{n \to \infty}\sqrt{\pi}=\sqrt{\pi}$$

If someone can help me and tell me where is my mistake, I will appreciate it very much.

I think my mistake is because I just found that it approximated it but not equal.

Answer 1

The main problem I see is that in using Stirling's approximation for both factorials you haven't proven that the error in the approximation disappears in the limit. It may happen to give the right answer in this case, but it is quite easy to construct situations where blindly applying approximations doesn't work - for example, this video where applying $\sin x \sim x$ in the limit $\lim_{x \rightarrow 0} \frac{1}{\sin^2 x} - \frac{1}{x^2}$ gives the wrong answer.

To approach the limit properly, you would need to break it into component limits whose values you can evaluate, which can still include a more careful application of Stirling's formula. For example, you could start something like this:

$$\lim_{n \rightarrow \infty} \frac{(n!)^2 2^{2n}}{(2n)!\sqrt{n}} = \lim_{n \rightarrow \infty} \left(\frac{n!}{\sqrt{2 \pi n} \left( \frac{n}{e} \right)^n} \right)^2 \left( \frac{\sqrt{2 \pi (2n)}\left( \frac{2n}{e} \right)^{2n}}{(2n)!} \right) \frac{2^{2n}}{\sqrt{n}} \frac{\left(\sqrt{2 \pi n} \left(\frac{n}{e} \right)^n \right)^2}{\sqrt{2 \pi (2n)} \left(\frac{2n}{e} \right)^{2n}}$$

where all I've done is "multiply by 1" (i.e. introduce the same factor into both the numerator and denominator) and broken things up into three terms. At this point you can then note that the limits of the first two terms are both 1 thanks to Stirling's formula, and the third term is essentially what you got in your solution, and so you can apply the limit rules to split this into the product of three separate limits and finish it off.

Answer 2

When you face problem with factorials, in my humble opinion, the easiest it to switch to the gamma function (which is continuous), then to tage logarithms every time it is possible) and only then to use Stirling approximation.

Whet is also good is, if you can, to take more terms that those required for the limit to have good approximations.

Let me take your problem $$A_n=\frac{(n!)^2\,2^{2n}}{(2n)!\,\sqrt{n}}=\frac{2^{2n}\,\Gamma (n+1)^2}{\Gamma (2 n+1)\, \sqrt n}$$ $$\log(A_n)=2 n \log (2)-\frac{\log (n)}{2}+2 \log (\Gamma (n+1))-\log (\Gamma (2 n+1)) $$

Now, apply twice Stirling approximation and continue with Taylor series to obtain $$\log(A_n)=\frac{1}{2}\log (\pi )+\frac{1}{8 n}-\frac{1}{192 n^3}+O\left(\frac{1}{n^5}\right)$$

Now, use $$A_n=e^{\log(A_n)}=\sqrt{\pi }\Bigg(1+\frac{1}{8 n}+\frac{1}{128 n^2}+O\left(\frac{1}{n^3}\right) \Bigg)$$

This gives you

• the limit that you properly found
• how the limit is approached (from above)
• a good approximation (the relative error is less than $0.01$% as soon as $n >3$)

Now, suppose that you look for a real $n$ such that $A_n=1.8$. Solving the quadratic leads to $$n=\frac{\sqrt{5 \left(18 \sqrt{\pi }-5 \pi \right)}+5 \sqrt{\pi }}{16\left(9-5 \sqrt{\pi }\right)}=\color{red}{8.10}513$$ while the "exact" solution (given using Newton method) is $8.10033$

And, be sure that we could do much better.

Answer 3

The solution is not quite standard. I want to show some beauty. In this solution I will use the Wallis integral $$W_{2n}=\int\limits_{0}^{\pi/2}\sin^{2n}xdx\Leftrightarrow \frac{2}{\pi}W_{2n}=\frac{2}{\pi}\int\limits_{0}^{\pi/2}\sin^{2n}xdx=\frac{(2n)!}{2^{2n}(n!)^2}$$ $$W_n\sim \sqrt{\frac{\pi}{2n}}\Rightarrow \frac{(2n)!\sqrt{n}}{2^{2n}(n!)^2}=\frac{\sqrt{2}}{\pi}\sqrt{2n}W_{2n}\sim \frac{\sqrt{2}}{\pi}\cdot \sqrt{\frac{\pi}{2}}=\frac{\sqrt{\pi}}{\pi}\Rightarrow \lim_{n\to\infty}\frac{\left(n!\right)^{2}2^{2n}}{\left(2n\right)!\sqrt{n}}=\sqrt{\pi}$$