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quasi
Posts:
9,080
Registered:
7/15/05
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Re: Product with elementary methods
Posted:
Jan 15, 2013 1:37 AM
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Ignacio Larrosa Cañestro 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.
Let
f(n) = 1 + 1/n^3
g(n) = 1 + 1/n
h(n) = 1 - 1/n + 1/n^2
k(n) = 1 - 1/n
Identically, f(n) = g(n)*h(n).
Also,
h(n) <= k(n+2)
<=> 1 - 1/n + 1/n^2 <= 1 - 1/(n+2)
<=> 1/(n+2) <= 1/n - 1/n^2
<=> n^2 <= n^2 + n - 2
<=> n >= 2
Let
x = Prod(f(n), n = 1..2013)
a = Prod(g(n), n = 1..2013)
b = Prod(h(n), n = 2..2013)
c = Prod(k(n), n = 4..2015)
Our goal is to show x < 3.
Now
a = (1 + 1/1)*(1 + 1/2)*(1 + 1/3) ... (1 + 1/2013)
= (2/1)*(3/2)*(4/3)* ... * (2014/2013)
= 2014
and
c = (1 - 1/4)*(1 - 1/5)*(1 - 1/6) ... (1 - 1/2015)
= (3/4)*(4/5)*(5/6)* ... * (2014/2015)
= (3/2015)
Then
x = a*h(1)*b = a*b < a*c = 3*(2014/2015) < 3
as required.
quasi
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