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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 ?

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2 Answers 2

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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.

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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 $$\lim_{n\to \infty} \sqrt[n] \frac{(2n)!}{(n !)^2}=4.$$

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  • $\begingroup$ Ok. thanks a lot. $\endgroup$
    – xpaul
    Jul 5 at 10:17

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