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.

(In reply to re(2): possible method. by Gamer)

The way Sofia did it:

Let p'=2/3* 4/5..    *96/97*98/99*1 .

Since 2/3>1/2,  4/5>3/4    ...   98/99>97/98   and 1>99/100

and      p*p'= 1/100= (.1)*(.1)     

p'>p

  SO  p   IS LESS YHAN 1/10.

Posted by Ady TZIDON on 2010-12-14 07:51:04