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 Hail to thee, Sofia! (Posted on 2010-12-13)

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.

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 re: possible method. | Comment 5 of 10 |
(In reply to possible method. by broll)

Just to tidy up a couple of things:

1/100 = (1/2*3/4*5/6...)*(2/3*4/5*5/6*7/8...)

should be

1/100 = (1/2*3/4*5/6...)*(2/3*4/5*6/7*8/9...)

More substantively, if written

1/100 = (1/2*3/4*5/6...*99/100)*(2/3*4/5*6/7*8/9...*98/99)

it's seen that the right-hand parenthetical factor has one fewer internal fraction factors than the left-hand parenthetical factor, and so not only is each factor larger reading from left to right, but the loss of one factor entirely also enlarges the overall factor since the individual factors are fractions under 1.

I only bring it up since, if the count of factors were the other way around, the argument presented would not be valid, so this is an essential part of the proof.

 Posted by Charlie on 2010-12-14 02:54:20

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