Let $a_n$ be a monotonic and bounded sequence, WLOG let assume it is monotonic increasing.
$a_n$ is bounded therefore there is a Supremum, $Sup(a_n)=a$, therefore $a_n<a+\epsilon$.
On the other hand due to $Sup(a_n)=a$, there is $N$ such that $a-\epsilon<a_N\leq a_n$ and together $a-\epsilon<a_N\leq a_n<a+\epsilon$
therefore $a-\epsilon< a_n<a+\epsilon$ and $lim_{n\to \infty}a_n=a$
Is the proof valid? does it apply to strictly monotonic sequence too?
Best Answer
Look good, you showed the monotonic increasing case converges to the least upper bound which is a, which is correct. For the decreasing case it should converge to the greatest lower bound (the inf(an) ). But I think it is good enough to show the increasing case and then say a similar proof follows for the decreasing case. Or you could just use the negative numbers in the increasing case and that would be a decreasing sequence that converges to the greatest lower bound. Yes it applies to the strict case as well. Since a strictly increasing or decreasing monotonic sequence is well increasing or decreasing.