1
$\begingroup$

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

$\endgroup$
8
$\begingroup$

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, $A_n\to +\infty$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.