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The limit:$$\lim_{n\to \infty}\left(\dfrac{\binom{3n}{n}}{\binom{2n}{n}}\right)^\frac{1}{n}$$

What I did was put limit = $L$. Then,

$$\begin{align}\log(L)&={\lim_{n \to \infty}}\dfrac{1}{n}\cdot\sum_{r=0}^{{n-1}} \log\left(\dfrac{3-\frac{r}{n}}{2-\frac{r}{n}}\right)\\ &=\int_0^1 \log\left(\dfrac{3-x}{2-x}\right)dx\\ &=\log\left(\dfrac{27}{16}\right) \end{align}$$

Is this aproach correct? Is there other method.

Edit: I have corrected the expression for the limit.

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    $\begingroup$ There are mistakes, but the method is basically correct. First, the log of L is equal to the limit of the RHS (and you need not assume L exists, by the way). Also, the upper bound of the sum is $n-1$, not $n$. $\endgroup$ Apr 12, 2017 at 6:38
  • $\begingroup$ Hmm very true!! $\endgroup$
    – jonsno
    Apr 12, 2017 at 6:39
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    $\begingroup$ Another method is to use Stirling's formula. $\endgroup$ Apr 12, 2017 at 6:40
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    $\begingroup$ A very quick way to show that the limit exists, though not of finding its exact value: using that $$\left(\frac{n}{k} \right)^k \leq \binom{n}{k} \leq \left(\frac{en}{k}\right)^k$$ the limit expression is between $\left(\frac{3^n}{(2e)^n}\right)^{1/n}$ and $\left(\frac{(3e)^n}{2^n}\right)^{1/n}$; i.e. between $\frac{3}{2e}$ and $\frac{3e}{2}$, or $0.55$ and $4.08$. $\endgroup$ Apr 12, 2017 at 6:55
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    $\begingroup$ Thank you @OlivierOloa that solved my problem! $\endgroup$
    – jonsno
    Apr 15, 2017 at 12:16

1 Answer 1

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$$ \begin{aligned} \lim _{n\to \infty }\left(\frac{\binom{3n}{n}}{\binom{2n}{n}}\right)^{\frac{1}{n}} & = \lim _{n\to \infty }\left(\frac{n!\left(3n\right)!}{\left(2n\right)!^2}\right)^{\frac{1}{n}} \\& = \lim _{n\to \infty }\exp\left(\frac{\ln\left(\frac{n!\left(3n\right)!}{\left(2n\right)!^2}\right)}{n}\right) \\& = \lim _{n\to \infty }\exp\left(\frac{\ln\left(\frac{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)\left(\sqrt{2\pi \left(3n\right)}\left(\frac{3n}{e}\right)^{3n}\right)}{\left(\sqrt{2\pi \left(2n\right)}\left(\frac{2n}{e}\right)^{2n}\right)^2}\right)}{n}\right) \\& = \lim _{n\to \infty }\exp\left(\frac{\ln\left(\frac{3^{\frac{6n+1}{2}}}{2^{4n+1}}\right)}{n}\right) \\& = \lim _{n\to \infty \:}\exp \left(\frac{\ln \left(\frac{e^{\left(\frac{6n+1}{2}\right)\ln3}}{e^{(4n+1)\ln2}}\right)}{n}\right) \\& = \lim _{n\to \infty }\exp \left(\ln \left(\frac{27}{16}\right)-\frac{\ln \left(\frac{4}{3}\right)}{2n}\right) \\& = \color{red}{\frac{27}{16}} \end{aligned} $$

Solved with Stirling approximation $$x! \approx \sqrt{2\pi x}\left(\frac{x}{e}\right)^x, \text{ for } x \to \infty$$

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