# Calculate $\lim \sqrt[n] \frac{(2n)!}{(n !)^2}.$ [duplicate]

I want to calculate $$\lim_{n\to \infty} \sqrt[n] \frac{(2n)!}{(n !)^2}$$

According to Wolfram alpha https://www.wolframalpha.com/input?i=lim+%5B%282n%29%21%2F%7Bn%21%5E2%7D%5D%5E%7B1%2Fn%7D , this value is $$4$$, but I don't know why.

I have $$\sqrt[n]{\dfrac{(2n)!}{(n !)^2}}=\sqrt[n]{\dfrac{2n\cdot (2n-1)\cdot \cdots \cdot (n+2)\cdot (n+1)}{n!}}$$ but I have no idea from here.

Another idea is taking $$\log.$$

$$\log \sqrt[n] \frac{(2n)!}{(n !)^2}=\dfrac{\log \frac{(2n)!}{(n !)^2}}{n} =\dfrac{\log \dfrac{2n\cdot (2n-1)\cdot \cdots \cdot (n+2)\cdot (n+1)}{n!}}{n} =\dfrac{\log [2n\cdot (2n-1)\cdot \cdots \cdot (n+2)\cdot (n+1)]-\log n!}{n}$$.

This doesn't seem to work.

Do you have any idea or hint ?

• You're actually on the right track. Finish with this.
– J.G.
Jul 1 at 14:49
• Jul 5 at 6:05

Option $$1$$ : Use Stirling's Approximation to solve it easily

Option $$2$$:- You already have $$\dfrac{\log [2n\cdot (2n-1)\cdot \cdots \cdot (n+2)\cdot (n+1)]-\log n!}{n}$$

Well this is just :-

$$\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\log(\frac{n+r}{r})=\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\log(1+\frac{n}{r})=\int_{0}^{1}\log(1+\frac{1}{x})\,dx = \ln(4)$$ .

Option $$3$$. Use Cauchy's Limit Theorems which say that when you have a sequence $$\{x_{n}\}$$ whose limit exists finitely , then the Arithmetic mean of the first $$n$$ terms also converge to the same limit as $$n\to\infty$$. That is $$\displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\frac{x_{r}}{n}=\lim_{n\to\infty}x_{n}$$ . See here and here for example.

Thus the answer is $$e^{\ln(4)}=4$$

My Advice: Use Stirling's Approximation if you're allowed to use it because it is much quicker.

By the Stolz theorem, one has $$\begin{eqnarray} \lim_{n\to \infty} \ln\bigg[\sqrt[n] \frac{(2n)!}{(n !)^2}\bigg]&=&\lim_{n\to \infty} \frac{\ln(2n)!-2\ln(n !)}{n}\\ &=&\lim_{n\to \infty} \frac{\left[\ln(2n+2)!-2\ln((n+1)!)\right]-\left[\ln(2n)!-2\ln(n !)\right]}{(n+1)-n}\\ &=&\lim_{n\to \infty} \ln[(2n+2)(2n+1)]-2\ln(n+1)\\ &=&\lim_{n\to \infty} \ln\bigg[\frac{(2n+2)(2n+1)}{(n+1)^2}\bigg]\\ &=&\ln 4 \end{eqnarray}$$ and hence $$\lim_{n\to \infty} \sqrt[n] \frac{(2n)!}{(n !)^2}=4.$$

• Ok. thanks a lot. Jul 5 at 10:17