Evaluate $\lim_\limits{n \to \infty} \frac{((2n)!)^2}{2^{4n}(n!)^4}$

Question

$$\lim_\limits{n \to \infty} \frac{((2n)!)^2}{2^{4n}(n!)^4}=0$$

How can I show this limit converges to 0?

I tried to use Sandwich Theorem, but I can't find proper fuction.

Answer 1

Well, if you know:

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

So:

$$\frac{\left(\sqrt{2\pi\left(2\text{n}\right)}\left(\frac{2\text{n}}{e}\right)^{2\text{n}}\right)^2}{2^{4\text{n}}\left(\sqrt{2\pi\text{n}}\left(\frac{\text{n}}{e}\right)^\text{n}\right)^4}=\frac{1}{\text{n}\pi}\tag2$$

Answer 2

Comment: $$ \sum _{n=0}^{\infty } {\frac { \left( \left( 2\,n \right) ! \right) ^{ 2}{x}^{2\,n}}{ \left( n! \right) ^{4}}} =\;{}_2F_1\left(\frac12,\frac12;1;16x^2\right) =\frac{2}{\pi}K\left( 4\,x \right) $$ where $K$ is the complete elliptic integral of the first kind. If we happen to know that $K(z)$ has a logarithmic singularity at $z=1$, then we can conclude that your sequence goes to $0$. We could also get this from knowledge of the singularity of ${}_2F_1(\frac12,\frac12;1;z)$ at $z=1$. Gauss already did the asymptotics of ${}_2F_1$ at its nearest singularity.

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Here is a result of Gauss. Study of the hypergeometric differential equation shows that ${}_2F_1\left(a,b;c;z\right)$ is holomorphic on the open unit disk, and (unless the series terminates) has a singularity at $z=1$ but no other singularity in the closed unit disk.

Theorem 2.1.3 in

Andrews, George E.; Askey, Richard; Roy, Ranjan, Special functions, Encyclopedia of Mathematics and Its Applications. 71. Cambridge: Cambridge University Press. xvi, 664 p. (1999). ZBL0920.33001.

$$ {}_2F_1\left(a,b;a+b;z\right) \sim \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\;\log\left(\frac{1}{1-z}\right) \quad\text{as }z \to 1^- . \\\text{so}\\ {}_2F_1\left(\frac12,\frac12;1;z\right) \sim \frac{1}{\pi}\;\log\left(\frac{1}{1-z}\right) \quad\text{as }z \to 1^- . $$ But at $z=0$ we have the Maclaurin series $$ \log\left(\frac{1}{1-z}\right) = \sum_{n=1}^\infty \frac{1}{n}\;z^n \\ \text{so}\\ \frac{1}{\pi}\log\left(\frac{1}{1-z}\right) = \sum_{n=1}^\infty \frac{1}{\pi n}\;z^n $$

Next there is Darboux's formula. It lets us recover the asymptotic properties of the Maclaurin coefficients of a series from knowledge of the nearest singularity of the function in the complex plane.
Theorem 8.4 in

Gábor Szegö, Orthogonal Polynomials, American Mathematical Society, Colloquium Publications, Vol. XXIII. American Mathematical Society, Providence, R.I.

The result is that the Maclaurin coefficients $$ a_n := {\frac { \left( \left( 2\,n \right) ! \right) ^{ 2}}{ 16^n\;\left( n! \right) ^{4}}} $$ of ${}_2F_1\left(\frac12,\frac12;1;z\right)$ satisfy $$ a_n \sim \frac{1}{\pi n} \quad\text{as } n \to \infty . $$

Answer 3

Using this answer we have $$b_n=\frac{e^{2n}((2n)!)^2}{2^{4n}n^{2n}(n!)^2} \to 2$$ If $a_n$ is the sequence in question then we have $$\frac{a_n} {b_n} =\left(\frac{n^n} {e^n(n!)}\right)^2$$ and we can prove that $c_n=n^{n} /e^{n} n! \to 0$.

Let's observe that we have $$n\log\frac{c_{n+1}}{c_n}\to-\frac{1}{2}$$ (prove this using Taylor or LHospital). Hence there is a positive integer $m$ such that $$\log\frac{c_{n+1}}{c_n}<-\frac{1}{3n}$$ for all $n\geq m$. Adding such equations for all values of $n=m, m+1,\dots,N-1$ we get $$\log\frac{c_N}{c_m}<-\frac {1}{3}\sum_{n=m}^{N-1}\frac{1}{n}$$ The right hand side tends to $-\infty $ and therefore so does the left hand side as $N\to\infty $. It follows that $c_n\to 0$ as $n\to\infty $. Thus the desired limit in question is $0$.

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It turns out that the technique used for $c_n$ can be used directly for $a_n$. We have $$\frac {a_{n+1}}{a_n}=\frac{(2n+1)^2} {(2n+2)^2}\to 1$$ and hence $$n\log\frac{a_{n+1}}{a_n}=n\cdot\frac{\log(a_{n+1}/a_n)}{(a_{n+1}/a_n)-1}\cdot\left(\frac{a_{n+1}}{a_n}-1\right)\to - 1$$ and we can proceed in the same manner and show that $a_n\to 0$.

More generally this shows that if $n\left(\left|\dfrac{x_{n+1}}{x_n}\right|-1\right)$ tends to a negative number then the sequence $x_n\to 0$. If this limit is less than $-1$ then the series $\sum x_n$ is absolutely convergent (see Raabe's test).