Difference between revisions of "1970 IMO Problems/Problem 3"
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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. | 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== | ==See also== | ||
Revision as of 10:22, 11 August 2008
Problem
The real numbers 
satisfy the condition:
.
The numbers 
are defined by
(a) Prove that 
for all 
.
(b) given 
with 
, prove that there exist numbers 
with the above properties such that 
for large enough .
Solution
Contradiction to (a): Let 
. Thus 
and that sum tends to infinity as 
tends to infinity.
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 | ||
