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

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==Solution==
 
==Solution==
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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.
  
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==See also==
 
{{IMO box|year=1970|num-b=2|num-a=4}}
 
{{IMO box|year=1970|num-b=2|num-a=4}}
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]

Revision as of 10:21, 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

$b_n = \sum_{k=1}^{n} \left( 1 - \frac{a_{k-1}}{a_{k}} \right)$

(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

Contradiction to (a): Let $a_{k}=k$. Thus $b_n = \sum_{k=1}^{n} \left( 1 - \frac{k-1}{k} \right)=\sum_{k=1}^{n} \frac{1}{k}$ and that sum tends to infinity as $k$ tends to infinity.

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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