intereting one(inequality)

let\ S=\frac{1}{2}.\frac{3}{4}. \frac{5}{6}.\frac{7}{8}...\frac{99}{100}\\ \\ we\ have, \\ \\ \frac{5}{6}>\frac{4}{5},\\ \\ \frac{7}{8}>\frac{6}{7}\\ \\ .\\ .\\ .\\ \\ \frac{99}{100}>\frac{98}{99}\\ \\ multiplying\ all\\ \\ \frac{5}{6}.\frac{7}{8}...\frac{99}{100}>\frac{4}{5}.\frac{6}{7}.\frac{8}{9}...\frac{98}{99}\\ \\ multiplying\ both\ side\ by (\frac{1}{2}.\frac{3}{4})\\ \\ S>\frac{3}{8}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}...\frac{98}{99}\\ \\multiplying\ both\ sides\ by\ S\\ \\ S^2>(\frac{3}{8})^2.(\frac{4}{5}.\frac{6}{7}.\frac{8}{9}...\frac{98}{99}).(\frac{5}{6}.\frac{7}{8}...\frac{99}{100})\\ \\ S^2>(\frac{3}{40})^2\\ \\ => S>\frac{3}{40}>\frac{1}{15}