Difference between revisions of "1991 IMO Problems/Problem 6"
(created page; gave one argument and linked to second argument by Osmun Nal) |
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Revision as of 17:13, 21 November 2018
Problem
An infinite sequence of real numbers is said to be bounded if there is a constant
such that
for every
.
Given any real number , construct a bounded infinite sequence
such that
for every pair of distinct nonnegative integers
.
Solution 1
Since , the series
is convergent; let
be the sum of this convergent series. Let
be the interval
(or any bounded subset of measure
).
Suppose that we have chosen points satisfying
for all distinct . We show that we can choose
such that
holds for all distinct
. The only new cases are when one number (WLOG
) is equal to
, so we must guarantee that
for all
.
Let be the interval
, of length
. The points that are valid choices for
are precisely the points of
, so we must show that this set is nonempty. The total length
is at most the sum of the lengths
. This is
.
Therefore the total measure of is
, so
has positive measure and thus is nonempty. Choosing any
and continuing by induction constructs the desired sequence.
Solution 2
The argument above would not work for , since
only converges for
. But Osmun Nal argues in this video that
satisfies the stronger inequality
for all distinct
; in other words, this sequence simultaneously solves the problem for all
simultaneously.