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Prove that $$\lim_{n\to\infty}{\root{n}\of{\frac{(2n)!}{(n!)^2}}}=+\infty$$

(Edit: as noted in the answers the actual limit is 4; I stand corrected.)

The argument of the $n$th root goes to infinity since it is larger than $n$, also it grows faster than than $n^k$ for any $k$... but taking the $n$th root of any of these would make the limit 1.

Is there a simple argument or should I use some trick...?

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3 Answers 3

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Set $a_n=\binom{2n}{n}$. Then if the quotient sequence $\frac{a_{n+1}}{a_n}$ converges to a limit $L$, the root sequence $\sqrt[n]{a_n}$ has the same limit.

As $$ \frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{(n+1)^2}=4·\frac{1+\frac1{2n}}{1+\frac1n} $$ does indeed converge, both sequences have the same limit $L=4$.


Bonus info: See also Newton's binomial series where $$ (1-x)^{-\frac12} =\sum_{k=0}^\infty\binom{-\frac12}k· (-x)^k =\sum_{k=0}^\infty\binom{2k}{k}·\left(\frac x4\right)^k $$ which has, as all binomial series, $R=1$ as radius of convergence, giving also the limit of the root sequence as $4$ by the theorem of Cauchy-Hadamard on the sharpness of the root formula for that radius.

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  • $\begingroup$ The first part seems to answer my question... I forgot about the quotient rule. Thanks! But I don't follow the second part... What is the definition of combinations{-1/2}{k} with -1/2 on the top? And how did you get from that to the right hand side (which includes $n$ - should that be $k$?) $\endgroup$ Dec 3, 2016 at 13:12
  • $\begingroup$ You are right, it should be $k$. $\binom{x}k=\frac{x(x-1)…(x-k+1)}{k!}$, so that $\binom{-1/2}k=(-\frac12)^k\frac{1·3·5···(2k-1)}{k!}$ and the fill the numerator using $(2j) = 2·j$ to get $(2k)!$ adding factors $2^k·k!$ to the denominator. $\endgroup$ Dec 3, 2016 at 13:31
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The limit is not infinite, it is $4$. This can be seen without the use of Stirling's formula. Knowing that $\binom{2n}{k} \le \binom{2n}{n}$ for $0 \le k \le 2n$, we get the following estimate from the binomial theorem:

$$\frac{4^n}{2n} = \frac{1}{2n}(1 + 1)^{2n} = \frac{1}{2n} \sum \limits_{k = 0}^{2n} \binom{2n}{k} \le \frac{1}{2n}\sum \limits_{k = 0}^{2n} \binom{2n}{n} = \binom{2n}{n} \le \sum \limits_{k = 0}^{2n} \binom{2n}{k} = (1 + 1)^{2n} = 4^n$$

Using the squeeze theorem we can conclude $\sqrt[n]{\frac{(2n)!}{n!^2}} = \binom{2n}{n}^{1/n} \xrightarrow{n \to \infty} 4$, since $\sqrt[n]{2n} \xrightarrow{n \to \infty} 1$.

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  • $\begingroup$ Using binomials is a neat trick. Thanks! $\endgroup$ Dec 3, 2016 at 15:28
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By using Stirling approximation, it turns out that the limit should be $4$.

Alternatively, you can also use these estimates $${4^n\over 2\sqrt{n}}\leq {2n\choose n}\leq{3\cdot 4^n\over4\sqrt{n+1}.}$$

See Elementary central binomial coefficient estimates

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  • $\begingroup$ Yep, also in my calculation.. $\endgroup$
    – Pieter21
    Dec 3, 2016 at 11:29

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