Indian Statistical Institute B.Math & B.Stat : Inequality

Prove that \[ \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} < \frac{1}{\sqrt{2n+1}} \] Let \(a,b > 0 \quad and \quad a < b \), we first show that \( \frac{a}{b} < \frac{a+1}{b+1}\). Now \( \frac{a+1}{b+1} - \frac{a}{b} = \frac{b-a}{b(b+1)} > 0 \quad since \quad b-a,a,b >0\). $$$$ Using the above result we see that \[\frac{1}{2} < \frac{2}{3} \] \[\frac{3}{4} < \frac{4}{5} \] \[\dots\] \[ \frac{2n-1}{2n} < \frac{2n}{2n+1} \] Let \( x_n = \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} \quad and \quad y_n = \frac{2}{3}. \frac{4}{5}.\frac{6}{7}. \frac{8}{9} \dots \frac{2n}{2n+1} \). Using the above result it is easy to see that \( x_n < y_n \). $$$$ Now \[ x_n^2 = x_n . x_n < x_n .y_n = \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} . \frac{2}{3}. \frac{4}{5}.\frac{6}{7}. \frac{8}{9} \dots \frac{2n}{2n+1} = \frac{1}{2n+1} \] \[ \implies x_n < \frac{1}{\sqrt{2n+1}} \quad i.e., \quad \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} < \frac{1}{\sqrt{2n+1}}\]