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 Over fourteen (Posted on 2011-12-02)
The sequence a(1),a(2),a(3),..., is formed according to the recursive rule a(1)=1, a(2)=a(1)+1/a(1),..., a(n+1)=a(n)+1/a(n), ...
Prove that a(100) > 14.

No direct evaluation, of course.

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 Solution(somewhat messy :D) | Comment 2 of 4 |
first of all lets reform:

a(n+1)=a(n)+1/a(n1)→
a(n+1) = a(n)*(1+1/a(n)²)

we compute the second term
a(2) = 1+1/1 = 2

using the reformed formula we conclude that if a(100) is to be below or equal to 14 then a(100)/a(2) <= 7
that means that in the reformed formula the term 1/a(n)² would be at the worst case 1/7².

so if (1+1/49)^99 > 7 then n(100) > 14

(1+1/49) ^99 > 7→
(50/49)^99 > 7→
50^99 > 7^199
(7²+1)^99 > 7^199

if we expand the left side we have:

7^198+99*7^197+....

we have that 99 > 7^2 so
99*7^197 > 7^199

(1+1/49)^99 > 7

so this fact contradicts that n(100) can be below or equal to 14.

 Posted by John Dounis on 2011-12-06 13:24:30

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