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

 Messages: [ Previous | Next ]
 quasi Posts: 12,067 Registered: 7/15/05
Re: Product with elementary methods
Posted: Jan 15, 2013 1:37 AM

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

Date Subject Author
1/14/13 Ignacio Larrosa Cañestro
1/15/13 quasi
1/20/13 Michael Press