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,



What is required for a monotone increasing sequence to converge?

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, $A_n\to +\infty$.

  • $\begingroup$ Why $\exists M$ and not $\forall M$? I'd have bet it should've been $\forall M$ but since you are two to use $\exists M$, there must be something I missed... $\endgroup$ – xavierm02 Jan 12 '13 at 21:59
  • $\begingroup$ I still don't get it... "has no upper bound" $\equiv$ $\lnot$("has upper bound")$\equiv \lnot(\exists M, \forall N, A_N < M) \equiv \forall M \lnot(\forall N, A_N < M)\equiv \forall M ,\exists N, \lnot(A_N<M)\equiv \forall M ,\exists N, A_N\ge M$ $\endgroup$ – xavierm02 Jan 12 '13 at 22:08
  • $\begingroup$ i clicked it twice, my bad, forgive my noobness $\endgroup$ – user5208 Jan 13 '13 at 0:58
  • $\begingroup$ it was so right, i clicked it twice hahahaha $\endgroup$ – user5208 Jan 13 '13 at 1:01

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


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