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

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.

@@ Line 17: / Line 17: @@
 ==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.
+==See also==
 {{IMO box|year=1970|num-b=2|num-a=4}}
 [[Category:Olympiad Number Theory Problems]]

1970 IMO (Problems) • Resources
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Difference between revisions of "1970 IMO Problems/Problem 3"

Revision as of 10:21, 11 August 2008

Problem

Solution

See also