$\lim_{n\to \infty}{2n \choose n}^{\frac{1}{2n}} = 2$ according to wolframalpha.
Does anyone see how to get this limit of $2$? Is it a numerical estimate or analytically exact?
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Alternative computation here: http://math.stackexchange.com/questions/701406/evaluating-lim-n-to-infty-frac12n-log-left2n-choose-n-right/701426#701426 – heropup Apr 15 '14 at 15:53
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Since $\binom{2n}{n}$ is the largest element among all $\binom{2n}{k}$ and $\sum\limits_{k=0}^{2n} \binom{2n}{k} = 2^{2n}$, we have $$\frac{2^{2n}}{2n+1} \le \binom{2n}{n} \le 2^{2n} \quad\implies\quad 2 \left(\frac{1}{2n+1}\right)^{1/2n} \le \binom{2n}{n}^{1/2n} \le 2$$ Since $\left(\frac{1}{2n+1}\right)^{1/2n} \to 1$ as $n \to \infty$, we have $\binom{2n}{n}^{1/2n} \to 2$ as $n \to \infty$.
achille hui
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@schulstud you are right. Strictly speaking, it is $<$ on both sides for finite $n \ge 1$. For simplicity, I edit one of them to $\le$ but forget the other ;-p – achille hui Apr 15 '14 at 16:46
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It follows from Stirling's formula, which implies the well-known asymptotic for central binomial coefficients $\binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}}$. Then
$$\binom{2n}{n}^{\frac{1}{2n}} \sim \left(\frac{4^n}{\sqrt{\pi n}}\right)^{1/(2n)} = 2\cdot (\pi n)^{-\frac{1}{4n}}\rightarrow 2$$
as $n\rightarrow\infty$.
mathse
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