Let $(k,+,.,0,1,<_k)$ be an ordered field, $| . |_k = x \mapsto \max(x,-x)$.
If $\lambda$ is an ordinal, you can define convergent $\lambda$-sequences: maps $f: \lambda \rightarrow k$ satisfying $\forall \varepsilon >_k 0(\exists \alpha \in \lambda(\forall \beta \ni \alpha(|f(\beta) - x|_k <_k \varepsilon)))$ for some $x \in k$. (you can also define Cauchy $\lambda$-sequences in a similar way)
If $cf(k)$ is the least order type of cofinal well ordered subsets of $k$ (or equivalently the least cardinal of cofinal subsets of $k$), then you get the nice property that $cf(k)$ is the least ordinal such that every bounded monotone $cf(k)$-sequence is a Cauchy $cf(k)$-sequence whenever cf(k) $> \aleph_0$. I can't find a way to do without this hypothesis but my method might not be the right one.
edit: This is actually false (see Eric Wofsey's answer).
This means that $cf(k)$ is the least ordinal such that $cf(k)$-sequentially closed sets are closed sets with respect to the order topology. Also, $k$ is complete (there is no dense embedding of $k$ into a proper linearly ordered field) iff its Cauchy $cf(k)$-sequences are convergent, but this result I can prove without assuming that $cf(k)$ is uncountable.
In the case when $cf(k) = \omega_0$, proving that bounded monotone sequences are Cauchy sequences is the same as proving that $k$ is archimedian. So my question is: are there non archimedian linearly ordered fields with countable cofinality? Do we even know the cofinality of some famous non archimedian fields such as fields of hyperreals? (without assuming CH)
There are lots of non-archimedean ordered fields of countable cofinality. For instance, if $k_0$ is any archimedean field, you can take $k=k_0(x)$ where $x$ is greater than every element of $k_0$ (the powers of $x$ are cofinal).
I also believe your result is incorrect even for fields of uncountable cofinality. For instance, let $\lambda$ be an uncountable regular cardinal and let $k=\mathbb{Q}(x_\alpha)_{\alpha<\lambda+\lambda}$, ordered such that $x_\alpha$ is greater than every element of $\mathbb{Q}(x_\beta)_{\beta<\alpha}$. Then $k$ has cofinality $\lambda$ (the sequence of $x_{\lambda+\alpha}$ for $\alpha<\lambda$ is cofinal), but the sequence $(x_\alpha)_{\alpha<\lambda}$ is a bounded increasing sequence of length $\lambda$ that is not Cauchy.