## A Low Bound for 1/2 · 3/4 · 5/6 · ... · (2n-1)/2n

Elsewhere we established an inequality

$\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{2n-1}{2n} \lt \frac{1}{\sqrt{3n+1}}$

which appeared an easier target for the mathematical induction than its weakened variant

$\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{2n-1}{2n} \lt \frac{1}{\sqrt{3n}}$

In an early issues of the Russian Kvant magazine (Kvant, n 5, 1970, M23), the famous mathematician A. O. Gelfond offered a two-sided inequality: for $n \gt 1$

$\displaystyle \frac{1}{2\sqrt{n}} \lt \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{2n-1}{2n} \lt \frac{\sqrt{3}}{2\sqrt{2n}}$

The right inequality is actually weaker than the one already established. The left inequality can be proved by the mathematical induction, but this is not how Gelfond chose to do that.

To remind, we wish to prove the low bound

$\displaystyle \frac{1}{2\sqrt{n}} \lt \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{2n-1}{2n}$

Let $A(n)$ be the right-hand side of the inequality and $B(n)$ its left-hand side. Thus we need to prove that $B(n) \lt A(n)$, $n \gt 1$. This result follows by an inductive argument from

$\displaystyle B(n+1) \lt B(n)\cdot \frac{2n + 1}{2(n+1)} \lt A(n)\cdot \frac{2n + 1}{2(n+1)} = A(n+1).$

What it takes is to verify that

$\displaystyle \frac{n}{n+1}=\bigg[\frac{\sqrt{n}}{\sqrt{n+1}}\bigg]^{2}\lt \bigg[\frac{2n + 1}{2(n+1)}\bigg]^{2}$

Gelfond himself solved the problem directly:

\begin{align} A(n)^2 & = \frac{1^2}{2^2}\cdot\frac{3^2}{4^2}\cdot\frac{5^2}{6^2}\cdot\ldots\cdot\frac{(2n-1)^2}{(2n)^2} \\ & = \frac{1}{2}\cdot\frac{3^2}{2\cdot 4}\cdot\frac{5^2}{4\cdot 6}\cdot\ldots\cdot\frac{(2n-1)^2}{(2n-2)2n}\cdot\frac{1}{2n}. \end{align}

But since, for any integer $k \gt 1$,

$\displaystyle \frac{(2k-1)^2}{(2k-2)2k} = \frac{(2k-1)^2}{(2k-1)^2 - 1} \gt 1,$

then

$\displaystyle A(n)^2 \gt \frac{1}{2}\cdot\frac{1}{2n}.$

In other words,

$\displaystyle A(n) \gt \frac{1}{2\sqrt{n}}.$