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Re: Product with elementary methods
Posted:
Jan 20, 2013 3:01 PM


In article <kd1emb$gra$1@ignaciolarrosa.eternalseptember.org>, Ignacio Larrosa Cañestro <ilarrosa@mundor.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/(n1)^2  1/n^2 = (2.n1)/(n(n1))^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



