Proof
Let
denote a bound from above, so that
holds for all . We assume that is not a Cauchy sequence. Then there exists some
such that for every , there exist indices
fulfilling
.
Because of the monotonicity, there is also for every an
with
.
Hence, we can define inductively an increasing sequence of natural numbers satisfying
-
-
and so on. On the other hand, there exists, due to the
axiom of Archimedes,
some
with
-
The sum of the first differences of the
subsequence
, ,
is
This implies
,
contradicting the condition that is an upper bound for the sequence.