Thread: Sequence convergence

Not sure how to go about this one.

If pn is a non-decreasing sequence and there is a point, p, to the right of each point of the sequence, then the sequence converges to some point.

Originally Posted by spikedpunch

If pn is a non-decreasing sequence and there is a point, p, to the right of each point of the sequence, then the sequence converges to some point.

That is an oddly stated problem.
Mathematically translated it says “A non-decreasing sequence bound above converges”.
Which of course is true. The proof follows from the completeness property.
The sequence converges to its least upper bound.

Originally Posted by spikedpunch

Not sure how to go about this one.

If pn is a non-decreasing sequence and there is a point, p, to the right of each point of the sequence, then the sequence converges to some point.

Isn't this the Monotone convergence theorem?

"If a sequence is monotone and bouned, then is converges"

here is an outline:

Suppose the sequence is increasing and bounded above.

Consider the set of all points in the sequece .

By the axiom of completeness, exists (since the set is a bounded subset of .

Assume , where .

Then, for , . Thus, there exists a point such that . Furthermore, because our sequence is increasing, it follows that, for every , . Thus, .

This implies

If I'm trying to prove that pn converges to some point, then I don't think I can assume that lim(pn)=s ......right?

Originally Posted by spikedpunch

If I'm trying to prove that pn converges to some point, then I don't think I can assume that lim(pn)=s ......right?

I really don't know how to answer that question.
Because I don't know what it means.
It is quite easy to prove this theorem: A bounded non-decreasing sequence converges to its least upper bound.

I'm saying that if I'm trying to prove a sequence converges, i can't assume that it has a limit and converges, because I would be assuming what I'm trying to prove.