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 $\lim_{n\rightarrow \infty}\frac{1}{n}=0=\inf\{s_n\}.$
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.
Perhaps just to expand on the commented answer, it is true that monotone sequences always converge to their infimum or supremum (depending of course on whether they are decreasing or increasing, respectively) assuming that such inf/sup exists. This fact is frequently called the Monotone Convergence Theorem.
However, just because a sequence is monotonic doesn't mean that it as an inf/sup. One very easy example is $\{s_n\} = n$. Of course this tends to $+\infty$, and so we could say that this sequence 'converges to $\infty$'. As long as you allow for this type of generalization to how inf/sup is defined (e.g. in this case we would say sup$\{s_n\} = \infty$), then yes, every monotonic sequence converges to its' infimum or supremum.
