All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 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.

 No Solution Yet Submitted by Ady TZIDON Rating: 3.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution(Correction) | Comment 3 of 4 |
I had a slight glitch in my previous post...

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)^98 > 7 then n(100) > 14

(1+1/49) ^98 > 7→
(50/49)^98 > 7→
50^98 > 7^197
(7²+1)^98 > 7^197

if we expand the left side we have:

7^196+99*7^195+....

we have that 99 > 7^2 so
99*7^195 > 7^197

(1+1/49)^98 > 7

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

 Posted by John Dounis on 2011-12-06 13:41:54
Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information