The title is the question.
I used the following definition of convergence to prove it:
$a_n$ converges to $x \in \mathbb R$ if $\forall \, \epsilon > 0, \, \exists \, N_0 \in \Bbb N$ s.t. $|a_n-x|<\epsilon, \,\forall \, n \geq N_0$
Definition of nondecreasing: $\forall \, n \in \Bbb N,\, a_{n+1} \geq a_n$.
Definition of bounded above: $\exists \, x \in \Bbb R$ s.t. $a_n \leq x \, \forall \, n \in \Bbb N$.
Proof: Set $\epsilon = |a_{n+1} - x|.$
$a_n \leq a_{n+1} => a_n - x \leq a_{n+1} - x => |a_n-x| \leq |a_{n+1} - x| => |a_n - x| \leq \epsilon.$
For $a_n<a_{n+1}$: The definition of convergence is satisfied.
$a_n = a_{n+1} \, \forall \, n \in \Bbb N => a_n = a_n \, \forall \, n \in \Bbb N => |a_n - x| = |a_n - a_n| = 0 < \epsilon, \, \forall \epsilon > 0.$