Simple math problems for recreation and research
"And yet the relation appears,
A small relation expanding like the shade
of a cloud on the sand,a shape on side of the hill"
-Wallace Stevens,'Connoisseur of Chaos' You shall find a [S] at the end of a post if solution can be found at one of its comments.Feel free to comment. HOME
Tuesday, January 23, 2007
An Inequality [S]
Prove that [1/2][3/4]......[9999/10000]< 1/100 [S] See comment for solution.I have solved using the technique that is called 'telescoping'.
I am not able to see your implication on line 5. Information 1: (lhs)^2 < 1/ 10001=0.(000999) Using calc, max value of lhs=0.009999 (correct of 6 decimals)
5 comments:
solution: observe that [n/(n+1)] < [(n+1)/(n+2)] for every natural n.
=>[1/2][3/4]...[9999/10000]<[2/3][4/5]...[10000/10001]
as each factor on lhs is less than corresponding term of rhs.
=>[lhs]^2 < [lhs][rhs]
lhs.rhs= [1/2][2/3]...[10000/10001]=1/10001
lhs=1/sqrt(10001)<1/100
[qed]
sqrt(a) > a If a < 1.
(i.e. sqrt(0.01) = 0.1)
Therefore
if lhs^2 < 1/10001
=> lhs > 1/100 .
What is the flaw in this argument ?
sqrt(10001)=100 + some +ve real num
=>sqrt(10001)>100
=>(1/sqrt(10001))<(1/100)
My point was taking square root of a number lesser than 1 increases the number.
So the inequality reverses.
I am not able to see your implication on line 5.
Information 1:
(lhs)^2 < 1/ 10001=0.(000999)
Using calc,
max value of lhs=0.009999
(correct of 6 decimals)
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