Difference between revisions of "1970 IMO Problems/Problem 3"

(I found a contradiction, please check.)
m (incorrect question)
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<math>b_n = \sum_{k=1}^{n} \left( 1 - \frac{a_{k-1}}{a_{k}} \right)</math>
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<math>b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}}</math>
 
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==Solution==
 
==Solution==
Contradiction to (a): Let <math>a_{k}=k</math>. Thus <math>b_n = \sum_{k=1}^{n} \left( 1 - \frac{k-1}{k} \right)=\sum_{k=1}^{n} \frac{1}{k}</math> and that sum tends to infinity as <math>k</math> tends to infinity.
 
 
 
{{solution}}
 
{{solution}}
  

Revision as of 13:03, 11 August 2008

Problem

The real numbers $a_0, a_1, \ldots, a_n, \ldots$ satisfy the condition:

$1 = a_{0} \leq a_{1} \leq \cdots \leq a_{n} \leq \cdots$.

The numbers $b_{1}, b_{2}, \ldots, b_n, \ldots$ are defined by

$b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}}$

(a) Prove that $0 \leq b_n < 2$ for all $n$.

(b) given $c$ with $0 \leq c < 2$, prove that there exist numbers $a_0, a_1, \ldots$ with the above properties such that $b_n > c$ for large enough $n$.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

1970 IMO (Problems) • Resources
Preceded by
Problem 2
1 2 3 4 5 6 Followed by
Problem 4
All IMO Problems and Solutions
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