Let p= 1/2*3/4*5/6*7/8……..*99/100.

Is p bigger than, equal to or less than 0.1 ?

The famous Sofia Kovalevskaya (1850-1891) solved this problem allegedly in no time, when she was only twelve years old.

Please provide your answer without programming and without direct evaluation.

It is pretty easy to rewrite the expression as
p = 100!/(2^100*(50!)^2)
then taking natural logarithm of both sides
ln p = ln 100! - 100ln 2 - 2 ln50!
and then using Stirlings Approximation
ln p ≈ 100ln 100 - 100 - 100ln2 - 2(50ln 50 - 50)
ln p ≈ 100ln 100 - 100ln 2 - 100ln 50
ln p ≈ 100(ln 100 - ln 2 - ln 50)
ln p ≈ 100(ln 100/(2*50))
ln p ≈ 100*0
ln p ≈ 0
p ≈ 1

This is clearly incorrect since p < 1/2

Edited on December 13, 2010, 5:43 pm

Posted by Jer on 2010-12-13 17:35:22