Bound in reflexive Banach space + monotonicity of sequence implies weak convergence of sequence? Let $V$ be a reflexive Banach space with a partial order relation $\leq.$ Furthermore, suppose it is a lattice. What further conditions does one need to have this property:

if $v_n$ is a bounded sequence in $V$ and $v_n \leq v_{n+1}$ for all $n$, then $v_n \rightharpoonup v$ in $V$.

Note the weak convergence is asked for for the original sequence (not a subsequence). Does anyone have a reference?

The property you are looking for is true in every reflexive ordered Banach space $V$ (provided that the cone $V_+$ is closed).
Proof. Let $(v_n)$ be an increasing norm bounded sequence in $V$.
Let us first note that if $(v_{n_k})$ and $(v_{m_k})$ are subsequences of $(v_n)$ and if they converge weakly to vectors $v$ and $w$, respectively, then $v = w$.
Indeed, let $v' \in V'$ be a positive and continuous linear functional on $V$. Both real-valued sequences
$$
  \big( \langle v', v_{n_k} \rangle \big)
  \quad \text{and} \quad 
  \big( \langle v', v_{m_k} \rangle \big)
$$
are subsequences of the increasing sequence $(\langle v', v_n \rangle)$, so their limits $\langle v', v \rangle$ and $\langle v', w \rangle$ coincides. Since the span of the dual cone is weak${}*$-dense in the dual space $V'$, we conclude that even $\langle w', v \rangle = \langle w', w \rangle$ for every functional $w' \in V'$, be it positive or not. Thus, $v = w$.
Now the claim easily follows: By the reflexivity, every subsequence of $(v_n)$ has a weakly convergent subsequence, and all these weakly convergent subsequences of subsequences have the same weak limit - say, $v$ - by what we have shown above. Thus, the sequence $(v_n)$ itself converges to $v$. $\square$
Remarks.
(1) I assumed that by a bounded sequence you mean a norm bounded sequence. If you mean order bounded instead, the situation changes considerably.
(2) The result and proof above do not assume that the ordered Banach space $V$ is lattice ordered.
