So the monotonic convergence theorem simply states that: $\{s_n\}$ is a monotonic sequence. Then
$\{s_n\}$ is convergent if and only if $\{s_n\}$ is bounded. by def this means $|s_n – L| < \epsilon$.
Furthermore it is true that
$s_1 < s_2 < …< s_n < … < s_m < ….< L$ because our sequence is monotonic and bounded, where $L$ is the limit and belong to $R$ and $n,m$ belong to the Natural Numbers and $n,m > n_0$ .
Then would the argument
$|s_n – s_m| < |s_n – L| < \epsilon$ (so the arbitrary distance from $s_n$ to $s_m$ is less than distance from $s_n$ to the limit is less than Epsilon)
then $|s_n – s_m| < \epsilon$ is also true, then by definition the question has been satisfied ?
Please excuse my English it is not my first language.
Best Answer
Assume $(s_n)$ increasing and bounded.
let $L=sup\{s_n,n\in\mathbb N\}$ and
$\epsilon>0$ given.
then
$\exists N\in \mathbb N \; \;$ $L-\epsilon<s_N\leq L$
but $(s_n)$ is increasing, thus
$\forall n>p>N \; $
$ L-\epsilon<s_N\leq s_p\leq s_n\leq L$.
finally
$\forall n>p>N \; \;\;|s_n-s_p|<\epsilon$
$(s_n)$ is then a Cauchy sequence.