To remind, we wish to prove the low bound

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

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 would follow immediately by an inductive argument from

\[ B(n+1) \lt B(n)\cdot \frac{2n - 1}{2n}. \]

Do verify, though, that the inequality does not hold, thus blocking the induction argument. Curiously, the induction applies to a stronger inequality:

\[ B(n-1) \lt A(n), n \gt 1. \]

Indeed, the inductive step is equivalent to proving

\[ B(n) \lt B(n-1)\cdot \frac{2n - 1}{2n}. \]

which, in turn, reduces to

\[ \frac{1}{\sqrt{2n}} \lt \frac{1}{\sqrt{2(n-1)}}\cdot\frac{2n-1}{2n}. \]

But this is equivalent to

\[ 2n \cdot (2n - 2) \lt (2n - 1)^2, \]

which is obviously true.

It is now an elementary matter to verify that \(B(2 - 1) \lt A(2)\) which sets the induction off.

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 \(k \gt 1\),

\[ \frac{(2k-1)^2}{(2k-2)2k} = \frac{(2k-1)^2}{(2k-1)^2 - 1} \gt 1, \]

then

\[ A(n)^2 \gt \frac{1}{2}\cdot\frac{1}{2n}. \]

In other words,

\[ A(n) \gt \frac{1}{2\sqrt{n}}. \]

Inequalities to prove: