Prove that $2^n(n!)^2 \leq (2n)!$

Question

Prove that $2^n(n!)^2 \leq (2n)!$

One can also use the following result to prove the above: $2 · 6 · 10 · 14 · · · · · (4n − 2) = \frac{(2n)!}{ n!}$.

The above relation gives, $(2n)!=2^n n! (1.3.5....(2n-1))$, how to go further.

Answer 1

You want to prove that $2^n (n!)^2 \le (2n)!$, but this is equivalent to:

$$2^n \le {2n \choose n} = \frac{(2n)!}{n! \cdot n!}$$

Suppose you have $n$ pairs, each consisting of one woman and one man. Right side counts the number of ways to choose $n$ people from that group. Left side counts the number of ways to choose one person from each pair. After noticing that combinatorial interpretation, the inequality becomes obvious.

Answer 2

$2^n(n!)^2\le (2n)!$. Now divide by $n!$ and get:

$$\iff \underbrace{2n(2(n-1))(2(n-2))\cdots(2(n-(n-1)))}_{n\text{ terms}}\le \underbrace{(2n)(2n-1)\cdots(2n-(n-1))}_{n\text{ terms}}$$

This is true because $2(n-k)\le 2n-k\iff k\ge 0$.

Answer 3

There are multiple ways. One can even use induction to prove it. Here's the easiest way:

$$2^n(n!)^2=(2^n\cdot n!)\cdot n!=(2n)!!\cdot n!\le (2n)!!\cdot (2n-1)!!=(2n)!$$

$$\therefore\quad 2^n(n!)^2\le (2n)!$$

Here, $n!!$ is the double factorial notation.

This is actually because $n!=1\cdot 2\cdot 3\cdots n\le 1\cdot 3\cdot 5\cdots (2n-1)=(2n-1)!!$

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Answer 4

As already noted by Leon Aragones, this is the same as proving $$ a_n=2^n \leq \binom{2n}{n}=b_n\tag{1}$$ that clearly holds for $n=1$. Since: $$ \frac{a_{n+1}}{a_n}=2,\qquad \frac{b_{n+1}}{b_n}=\frac{4n+2}{n+1}\geq 2 \tag{2}$$ we have that $(1)$ holds by induction.

Answer 5

It is sufficient to prove $$2^n\le\binom{2n}{n}\tag 1$$ First, note that $$2^n=(1+1)^n=\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}\tag 2$$

Second, note that $$\binom{2n}{n}={\binom{n}{0}}^2+{\binom{n}{1}}^2+\cdots+{\binom{n}{n}}^2\tag 3$$ From $(2),(3)$, you'll get $(1)$.

To get $(3)$, compare the coefficient of $x^n$ in the both sides of $$(1+x)^n(x+1)^n=(1+x)^{2n}.$$

Answer 6

Let $[k]$ denote the set $\{1,2,\ldots,k\}$. To see that $\binom{2n}n\ge2^n$ one can observe that $\binom{2n}n$ counts the number of subsets of size $n$ in $[2n]$, and $2^n$ counts the number of such subsets that have the form $S\cup\{n+i: i\in ([n]\setminus S)\}$. That is, subsets that contain exactly one of $i, i+n$ for each $i\in[n]$.

Answer 7

Another approach for this kind of problem : the use of Stirling approximation is very interesting $$m!\approx m^m \sqrt{2\pi m}\,e^{-m}$$ so $$\frac{(2n)!}{2^n (n!)^2}\approx \frac{2^n}{\sqrt{\pi } \sqrt{n}}$$ which is already greater than $1$ for $n=1$. Now, you can easily show (using its derivative) that the expression increases with $n$.