# An Inequality with Central Binomials

### Problem

Prove that for integer $n\ge 1,$

$\displaystyle \sqrt{2}\le\sqrt[n(n+1)]{{2\choose 1}{4\choose 2}\cdots{2k\choose k}\cdots{2n\choose n}}\lt 2.$

### Solution 1

Let

\begin{align} a&=2\cdot 4\cdot 6\cdot\ldots\cdot (2k)=(2k)!!\\ b&=1\cdot 3\cdot 5\cdot\ldots\cdot (2k-1)=(2k-1)!!\\ c&=1\cdot 2\cdot 3\cdot\ldots\cdot k=k! \end{align}

Clearly, $a\gt b\gt c,$ implying $a^2\gt ab\gt ac,$ meaning

$2^{k}\cdot (k!)^2\le (2k)!\gt 2^{2k}(k!)^2,$

or, $\displaystyle 2^{k}\lt \frac{(2k)!}{(k!)^2}\le 2^{2k}.$ Taking the product,

\displaystyle \begin{align} 2^{1+2+\ldots+n}\le{2\choose 1}{4\choose 2}\cdots{2k\choose k}\cdots{2n\choose n}\lt 2^{2(1+2+\ldots+n)} \end{align}

which is equivalent to the required inequality. For $n=1,$ $\sqrt{2}=\sqrt{2}\lt 2.$

### Solution 2

We prove that $2^k\le {2k\choose k} \lt 2^{2k}.$ Indeed, for the left inequality,

$\displaystyle {2k\choose k}=\sum_{i=0}^k{k\choose i}^2\ge\sum_{i=0}^k{k\choose i}=2^k.$

And, obviously, $\displaystyle {2k\choose k}\lt\sum_{i=0}^{2k}{2k\choose i}=2^{2k}.$ Taking the product we obtain the required inequality,

$\displaystyle \prod_{k=1}^{n}2^k\le\prod_{k=1}^{n}{2k\choose k}\le\prod_{k=1}^{n}2^{2k}$

where on the left, equality is attained for $n=1.$

### Solution 3

For the left inequality,

\displaystyle\begin{align} {2n\choose n}&=\frac{2n\cdot (2n-1)\cdot (2n-2)\cdot\ldots\cdot (n+2)\cdot (n+1)}{n\cdot (n-1)\cdot (n-2)\cdot\ldots\cdot 2\cdot 1}\\ &=\frac{2n}{n}\cdot\frac{2n-1}{n-1}\cdot\frac{2n-2}{n-2}\cdot\ldots\cdot\frac{n+2}{2}\cdot\frac{n+1}{1}\\ &\ge \underbrace{2\cdot 2\cdot 2\cdot \ldots\cdot 2\cdot 2}_{n\text{ times}}=2^n. \end{align}

The right-hand side is estimated via the binomial formula as in Solution 2.

### Solution 4

We prove by induction that $\displaystyle 2^k\le{2k\choose k}\lt 2^{2k}.$ Indeed, for $k=1,$ we have $2\le 2\lt 4.$ Further,

$\displaystyle {2(n+1)\choose n+1}=\frac{(2n+2)(2n+1)}{(n+1)^2}{2n\choose n}=\frac{2(2n+1)}{(n+1)}{2n\choose n}.$

But $\displaystyle 2\le \frac{2(2n+1)}{(n+1)}\lt 4.$

### Remark

A lower bound for the central binomial coefficient $\displaystyle {2n\choose n}$ is $\displaystyle \frac{4^n}{\sqrt{4n}}$ and so the limit should follow from Stirling's approximation. That it strictly increases to $2$ should follow by induction.

### Acknowledgment

The problem has been kindly posted by Dorin Marghidanu at the CutTheKnotMath facebook page, along with Solutions 1. Leo Giugiuc commented with Solution 2. Solution 3 is by Christopher D. Long; Solution 4 is by Sam Walters. Remark is due to Christopher D. Long.