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.

To start with, I totally misread the problem as (1/2*2/3*3/4*...99/100) which cancels to 1/100.

Then I thought:1/100 = (1/2*3/4*5/6...)*(2/3*4/5*6/7...) and either the evens or the odds must be bigger than 1/10 since their product is 1/100.

But 2/3>1/2,4/5>3/4 etc.

So the odds are less than the evens and hence less than 0.1

 

Edited on December 14, 2010, 2:59 am

Posted by broll on 2010-12-14 00:58:49