Are monotone limited hypersequences converging? Consider the usual ultrapower construction of the hyperreals $^*\mathbb R$.
Let $a:{}^*\mathbb N \to \mathbb L$ be an increasing limited hypersequence. Does $a$ have an upper bound in $\mathbb L$? 
Background:
It is well known that the classical porperty that any monotone bounded sequence converges does not carry over to the hyperreals since for example $a_n = n$ is bounded in $^*\mathbb R$ by any positive unlimited number, yet not converging since the is no least unlimited number.
However, I was wondering what happens if we look instead at genuine (not necessarily internal) hypersequences. The above question addresses this issue in weakend form. I noted that if we assume that $a$ is internal, then it is true since then $(a_n)_{n\in\mathbb N}$ is necessarily bounded in $\mathbb R$ $-$ else $a_N$ would be unlimited for some $N\in{}^*\mathbb N_\infty$  $-$ hence convergent. Thus the extended tail is in the halo of the limit and so also $(a_n)_{n\in{}^*\mathbb N}$ is bounded in $\mathbb L$.
In particular, if an increasing limited hypersequence $a:{}^*\mathbb N \to \mathbb L$  has an intenal majorant $^*b:{}^*\mathbb N \to \mathbb L, \forall n :b_n \ge a_n$ , then the answer is yes. However it seems very difficult to construct such a majorant since one need to ensure dominance for the whole extended tail.
I didn't had any luck yet finding a counterexample either, so any ideas or comment are greatly appreciated.
 A: Suppose $(a_n)$ is an increasing sequence (internal or external) without a real upper bound. For each $n\in\mathbb N$ choose $H=H(n)$ so that $a_H\geq n$. Define the set $A_n=\{H, H+1,H+2,\ldots\}\subseteq{}^\ast\mathbb N$.  Then $A_n$ is internal and nonempty. By choosing $H(n+1)\geq H(n)$ at each stage, we ensure that the sets $A_n$ form a nested decreasing sequence. By saturation, there is an element $K\in\cap_{n\in\mathbb N} A_n$. By construction $a_K> n$ for each $n\in\mathbb N$. Therefore $a_K$ is infinite, contradicting the hypothesis that each $a_n$ is finite. 
Therefore every such sequence $a_n$ is necessarily bounded.
A: $\DeclareMathOperator{\cof}{cof}$
Since the three posts are basically the same, I edited mine to set this in a general context. The following would require some work to properly use in your case but is otherwise elementary and generally useful:
Claim: If $X,Y$ are linearly ordered sets where $Y$ has no maximum, then there is an increasing cofinal map $f: X \rightarrow Y$ iff $\cof(X) = \cof(Y)^{\mathbf{hey}}$.
If $f$ is such a map, by the second caracterisation of cofinality there is a cofinal embedding $\varphi: \cof(X) \rightarrow X$. $f \circ \varphi$ is a cofinal increasing map $\cof(X) \rightarrow Y$, and it cannot be eventually constant for otherwise said constant would be the maximum of $Y$.
By $(ii)$, $\cof(X)$ is regular, which allows us to define inductively (and using $(i)$) a cofinal embedding $\psi: \cof(\alpha) \rightarrow \cof(X)$ such that $f \circ \varphi \circ \psi$ is strictly increasing (and cofinal), which implies by $(ii)$ that $\cof(Y) = \cof(X)$.
If $\cof(X) = \cof(Y) = 0$ then $X = Y =  \varnothing$, whence the map $\varnothing$ is solution.
If $0 <\cof(X) = \cof(Y)$, then fix cofinal embeddings $\varphi: \cof(X) \rightarrow X$ and $\psi: \cof(X) \rightarrow Y$ and define $f$ piecewise by $f(]-\infty;\varphi(0)[) = \{\psi(0)\}$ and $\forall \alpha < \cof(X), f([\varphi(\alpha);\varphi(\alpha+1)[) = \{\psi(\alpha)\}$ where $\varphi(\alpha+1) = +\infty$ if $\alpha+1 \notin \cof(X)$. $f$ is an increasing cofinal map $X \rightarrow Y$.
Fun fact, this implies that there being an increasing cofinal map between linear orders without maxima is an equivalence (pseudo-)relation.
[hey] The cofinality $\cof(X)$ of a linearly ordered set $X$ is  equivalently:
-$(i)$: the least cardinal of a cofinal subset of $X$
-$(ii)$: the least ordinal which embeds cofinaly in $X$
-$(iii)$: the unique regular ordinal which embeds cofinaly in $X$

Here $\cof(^*\mathbb{N}) = \cof(^*\mathbb{R})$ is uncountable by $\aleph_1$-saturation of $^*\mathbb{R}$ (or more simply by the lemma you proved in your own answer), while $\cof(\mathbb{L}) = \omega_0$, so any such sequence is bounded.
A: I actually had a similar idea as Mikhail myself:
Suppose that $a_n$ has no upper bound in $\mathbb L$. Then $A_n = \{N\in{}^*\mathbb N : a_N > n\}$ is non-empty for all $n$ and these sets form a decreasing chain. Choose a map $\psi:\mathbb N \to {}^*\mathbb N$ with $\psi(n)\in A_n$ for all $n$.
Lemma: Any map $\psi:\mathbb N \to {}^*\mathbb R$ is bounded.
Proof (via diagonalization): Let $\psi(n) = [w^n] = [(w^n_1,w^n_2,\ldots)]$. Define $\hat w_k = \max\{w^1_k,w^2_k,\ldots w_k^k\}$. Then for any fixed $n$, $\hat w_k \ge w^n_k$ for all but finitely many $k$. Hence $[\hat w] \ge [w^n]$ for all $n$ is an upper bound for $\psi$.
So there is an $M\in{}^*\mathbb N$ such that $a_M > a_{\psi(n)} > n$ for all $n$, contradicting the assumption that $a_n\in\mathbb L$ for all $n$. 
