Do bounded monotone sequences always converge to either its supremum or infimum?
For example, $\{s_n\}=\frac{1}{n}$
is a decreasingly monotonic sequence. Since $0\leq \frac{1}{n}\leq 1$ and $\inf\{s_n\}=0$ and $\sup\{s_n\}=1$, then
$\lim_{n\rightarrow \infty}\frac{1}{n}=0=\inf\{s_n\}.$
Best Answer
Yes. Let $a_n$ be a monotone increasing(the proof for monotone decreasing is analogous) sequence.
Let $a = sup_n \{a_n\}$. if $\{a_n\}$ is bounded than $a< \infty$ , but the claim remains true for $a=\infty$ (and you can think of the adjustments).
We want to show $lim_n a_n =a$.
So, Let $\epsilon >0$.
By definition of the supremum, there is $n_0$ s.t. $a_{n_0} > a -\epsilon$ thus $|a_{n_0}-a| =a-a_{n_0} <\epsilon$.
now, for all $n\ge n_0$ we get, since $a_n \ge a_{n_0} $ ,
$|a_n-a| = a-a_n\le a-a_{n_0} < \epsilon$.
So , $a_n \to a$ ,as we wanted.