1similarly we can proceed for the maximum value of S.
if nyone interested can try, or else will post the solution tomorrow,
prove that
\frac{1}{15}<\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}........\frac{99}{100}<\frac{1}{10}
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3 Answers
1let\ 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}
11the right hand side of the inequality can be proved as here http://targetiit.com/iit_jee_forum/posts/good_inequality_3921.html
