Limits of Monotonic Sequences
Appendices, Volume 3, Why Slopes and More Math.
Vocabulary.} A sequence $q_k$ with the property
$q_k\ge q_{k+1}$ for every whole number $k$, is said to be
a decreasing sequence. A sequence $q_k$ with the property
$q_k\le q_{k+1}$ for every whole number $k$, is said to be
an increasing sequence. A sequence $q_k$ is said to be
monotonic if it is an increasing sequence or a decreasing
sequence.
| [On Decreasing Sequences]
If the
infinite sequence $q_1,q_2,q_3\ldots$ has the property that
for each whole numbers $k$, $q_k\ge q_{k+1} \ge Q$ for some
real number $Q$, a lower bound, then the sequence converges to a limit
$A\ge Q$. The limit $A$ is the greatest lower bound of the
set of points in the sequence.
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Proof.
First, by mathematical induction, $nSecond, if there exist a whole number $K>0$ such that $k\ge K$
implies $q_k=q_K$ then the limit is given by $A=q_K$ and
the sequence takes on at most $K$ distinct values
$q_1,\ldots,q_K$. Otherwise for every $K>0$, there is a
$k>K$ such that $q_k\ne q_K$. In this case, the range of
values assumed by the sequence is an infinite set. The case
where the range of values is an infinite set is considered next.
An small assertion: observe $A > q_m$ for some $m$ implies
that $A > q_m\ge q_n$ for all $n \ge m$. This in turn
implies $A$ is not a limit point. The interval of length
$q_m-A$ centered at $A$ contains at most $m$ elements of
the sequence. That $A>q_m$ for some $m$ implies that
$A$ is not a limit point of the sequence $q_j\ge q_{j+1}$.
According to the Bolzano-Weierstrass theorem, the infinite
set formed by the sequence $q_j$ has a limit point $A$.
The contrapositive of the above assertion now
implies that the limit point $A\le q_p$ for all whole numbers $p$. But
now for each whole number $k$, the interval
$A+\frac1210^{-k}$ to $A$ (with $A$ included) contains
infinitely many elements of the sequence. Therefore there
is some sequence element $q_n$ with the property that
$A+\frac1210^{-k} \ge q_n \ge A$. Therefore $p \ge n$
implies $A+\frac1210^{-k} \ge q_n \ge q_p \ge A$ since
$q_p\ge A$ and $q_n\ge q_p$ whenever $p>n$.
Finally, since $k$ is arbitrary, we conclude include the
sequence $q_n$ converges to a limit $A$. Moreover, we can
say there are no points in the sequence $q_n$ to the left
of $A$.
The proof of the following theorem is similar.
| [On Increasing Sequences]
If the infinite
sequence $q_1,q_2,q_3\ldots$ has the property that for each
whole numbers $k$, $q_k\le q_{k+1} \le Q$ for some real
number $Q$, an upper bound, then the sequence converges to
a limit $B \le Q$. The limit $B$ is the least upper bound
of the set of points in the sequence.
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