2002 OIM Problems/Problem 5: Difference between revisions
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== Solution == | == Solution == | ||
{{ | Notice that every time we apply the recursion, we are essentially subtracting <math>\frac{1}{a_n}</math> from the current term. Clearly, the sequence is decreasing, so for all <math>n</math>, <math>a_n\le56</math>, so <math>\frac{1}{a_n}\ge\frac{1}{56}</math>. Then, after every application of the recursion, the value of <math>a_i</math> will decrease by at least <math>\frac{1}{56}</math>; clearly, it will eventually reach negative numbers. | ||
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406] | |||
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe18.htm | https://www.oma.org.ar/enunciados/ibe18.htm | ||
Latest revision as of 17:23, 5 May 2025
Problem
The sequence of real numbers 
is defined as:
![\[a_1 = 56, a_{n+1} = a_n - \frac{1}{a_n}\]](http://latex.artofproblemsolving.com/9/2/0/9201b9360f186cd4e092328a53a591f21f0c32cc.png)
for every integer 
.
Prove that there exists an integer 
, 
, such that 
.
~translated into English by Tomas Diaz. orders@tomasdiaz.com
Solution
Notice that every time we apply the recursion, we are essentially subtracting 
from the current term. Clearly, the sequence is decreasing, so for all 
, 
, so 
. Then, after every application of the recursion, the value of 
will decrease by at least 
; clearly, it will eventually reach negative numbers.
