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Topic: Product with elementary methods
Replies: 2   Last Post: Jan 20, 2013 3:01 PM

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Michael Press

Posts: 2,104
Registered: 12/26/06
Re: Product with elementary methods
Posted: Jan 20, 2013 3:01 PM
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In article <kd1emb$gra$1@ignaciolarrosa.eternal-september.org>,
Ignacio Larrosa Cañestro <ilarrosa@mundo-r.com> wrote:

> Prove that: Product (1 + 1 / n ^ 3), n, 1, 2013) <3
>
> Problem posed in the local phase of the Spanish Mathematical Olympiad,
> aimed at secondary school students (16/17 old)
> Must be resolved with elementary methods, appropriate to that level.


The product of the first two factors is 2(9/8) = 9/4.
Bound sum 1/n^3 with a telescoping sum.

1/(n-1)^2 - 1/n^2 = (2.n-1)/(n(n-1))^2
< 2/n^3

SUM 1/n^3 < (1/2)(1/4) = 1/8
3 <= n

Now apply the AGM inequality.

2011
( 1 1 )
PROD (1 + 1/n^3) <= ( 1 + ----- --- )
3 <= n <= 2013 ( 2011 8 )


( 1 1 )
<= ( 1 + --- + --- + ...)
( 8 64 )

9 8 72 72
PROD (1 + 1/n^3) <= -- --- = --- < --- = 3.
1 <= n <= 2013 4 7 28 24

--
Michael Press



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