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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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