Proof of the divergence of a monotonically increasing sequence Say that the limit of sequence $(A_n)$ as $n \to \infty$ equals $+\infty$ if for every $r \in \mathbb R$, there is an integer $N$ such that $(A_n) > R$ for all $n \geq N$.

Show that a divergent monotone increasing sequence converges to $+\infty$ in this sense.

I am having trouble understanding how to incorporate in my proof the fact that the sequence is monotonically increasing.
Any help would be appreciated,
Thanks
 A: What is required for a monotone increasing sequence to converge? 


*

*See Monotone Convergence Theorem for Sequences.$(*)$


By$\; (*):\;$ Since $A_n$ is a monotone increasing sequence, if $A_n$ were bounded above, then it would converge to some value $L < +\infty$. But we are given $A_n$ is a divergent monotone increasing sequence. Hence $A_n$ cannot be bounded above; i.e., $A_n$ has no upper bound. (That's simply applying the contrapositive of the monotone convergence theorem).
That is, there is no $M > 0$ such that $(A_n)$ is bounded above by $M$. This means
$$\forall M>0,\; \exists N\in \mathbb{N}:A_N>M.$$
And since we are given that $(A_n)$ is monotone increasing, $n\ge N\implies A_n>A_N>M$.
This holds for $M'<M$ too, so
$$\forall M>0,\;\;\exists N\in \mathbb{N}:\ n\ge N\implies A_n>M$$
and hence, $A_n\to +\infty$.
A: If $(a_n)$ is bounded above then by the Monotone Convergence theorem, $(a_n)$ converges which is a contradiction. Thus $(a_n)$ is not bounded above by some $M>0$, that is
$$\exists M>0\ \exists N\in \mathbb{N}:a_N>M$$
But as $(a_n)$ is increasing, 
$$n\ge N\implies a_n>a_N>M$$
This holds for $M'<M$ as well and so
$$\forall R>0\exists N\in \mathbb{N}:\ n\ge N\implies a_n>R$$
and so $a_n\to +\infty$.
Note that we only incorporate divergence to show un-boundedness. This is done by the Monotone Convergence Theorem for sequences
