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I've had a hard time computing the limit $\lim\limits_{n\to \infty }\frac {(2n)!}{(2^n(n)!)^2} $ either by bounding it or by simplifying it. I would appreciate some help.

(P.S. I came across this limit while using the ratio test to calculate the radius of convergence of the solution of $y''+xy'+2y=0 $ around $x_0=0$ in the form of power series.)

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    $\begingroup$ Stirling formula... $\endgroup$ – dan_fulea Feb 4 at 15:43
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An alternative to the use of Stirling's formula. \begin{align} \frac{(2\,n)!}{(2^n\, n!)^2}&=\frac{(2\,n-1)(2\,n-3)\dots5\cdot3\cdot1}{2^n\,n!}\\ &=\frac{2\,n-1}{2\,n}\cdot\frac{2\,n-3}{2\,n-2}\cdots\frac56\cdot\frac34\cdot\frac12. \end{align}

and now prove by induction that $$ \frac{2\,n-1}{2\,n}\cdot\frac{2\,n-3}{2\,n-2}\cdots\frac56\cdot\frac34\cdot\frac12\le\frac{1}{\sqrt{3\,n+1}}. $$ as in this question.

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As suggestes by dan_fulea within the comments we may use Stirling's formula in order to approximate the factorials. For sufficient large $n$ we got that

$$n!~\sim~\sqrt{2\pi n}\left(\frac ne\right)^n\tag1$$

Therefore your limit can be written as

\begin{align*} &\lim_{n\to\infty}\frac{(2n)!}{(2^n\cdot n!)^2}=\lim_{n\to\infty}\frac{\sqrt{2\pi (2n)}\left(\frac{2n}e\right)^{2n}}{2^{2n}\left(\sqrt{2\pi n}\left(\frac ne\right)^n\right)^2}=\frac1{\sqrt{\pi}}\lim_{n\to\infty}\frac1{\sqrt{n}~2^{2n}}\frac{e^{2n}}{e^{2n}}\frac{(2n)^{2n}}{n^{2n}}\\ =&\frac1{\sqrt{\pi}}\lim_{n\to\infty}\frac{2^{2n}}{\sqrt{n}~2^{2n}}=\frac1{\sqrt{\pi}}\lim_{n\to\infty}\frac1{\sqrt{n}}=0 \end{align*}

$$\therefore~\lim_{n\to\infty}\frac{(2n)!}{(2^n\cdot n!)^2}~=~0$$

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  • $\begingroup$ It seems the $2^n$ in the denominator is also supposed to be squared. $\endgroup$ – User8128 Feb 4 at 15:59
  • $\begingroup$ That’s weird because this thing 2n choose n over 4^n which is clearly less than 1 $\endgroup$ – HAMIDINE SOUMARE Feb 4 at 16:00
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Using Stirling's formula you get $$ \lim\frac{(2n)!}{(2^n n!)^2}= \lim\frac{\sqrt{4n\pi}(2n/e)^{2n}}{(2^n \sqrt{2n\pi} (n/e)^n)^2}= \lim \frac{1}{\sqrt{\pi n}}=0 $$

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Arithmetic inequality

$$\frac{2r-1+2r+1}{2}\geq \sqrt{(2r-1)(2r+1)}$$

$$2r\geq \sqrt{(2r-1)(2r+1)}$$

$$\frac{1}{2r}\leq \sqrt{\frac{1}{(2r-1)(2r+1)}}$$

$$0<\prod^{n}_{r=1}\frac{2r-1}{2r}\leq \prod^{n}_{r=1}\sqrt{\frac{2r-1}{2r+1}}=\frac{1}{\sqrt{2n+1}}\;\;, n\geq 1$$

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How about this solving it this way?

$\lim\limits_{n \to \infty} \frac{(2n)!}{{(2^nn!)}^2}=\lim\limits_{n \to \infty} \frac{{2n}\choose{n}}{{(2^n)}^2}=\lim\limits_{n \to \infty} \frac{\sum_{x=0}^{n}{{n \choose x} ^2}}{{(\sum_{x=0}^{n}{n \choose x}})^2}=0$

You can fill in the gaps (for example you'll have to mention the last step holds true because ${n \choose x} \geq 0$)

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