Here is the problem. How would I prove this problem?

http://i10.photobucket.com/albums/a1.../problem46.jpg

Printable View

- September 19th 2010, 11:38 AMPhyxius117Real Analysis (Sequences)
Here is the problem. How would I prove this problem?

http://i10.photobucket.com/albums/a1.../problem46.jpg - September 19th 2010, 11:48 AMPlato
Is it clear that

Can you finish? - September 19th 2010, 02:47 PMPhyxius117
Do I go with a direct proof?

like:

Given > 0

Then

Thus

but what do I do next? Do I split it up into two? - September 19th 2010, 03:17 PMPlato
- September 19th 2010, 03:24 PMPhyxius117
I don't quite understand. how did you get x_n > a/2 ?

- September 19th 2010, 03:28 PMPlato
- September 19th 2010, 04:59 PMPhyxius117
Thank you for all the help.

after looking at your hints on how to approach this problem, i'm trying hard to figure out how to solve it but I just don't understand what I am suppose to do. - September 20th 2010, 01:37 AMmagus
I happened on this problem and I was wondering if the following reasoning is enough to prove that .

If is monotonically increasing or non-decreasing then it must eventually pass in order to get to . If it is monotonically decreasing or non-incresing then all values of must be greater then . If it is none of those we can still choose a monotonic subsequence because all sequences have a monotonic subsequence and apply the logic previously stated.

Is this enough just for ? - September 20th 2010, 02:37 AMMathoMan
Hi. I like these assignments, they can be challenging. I am curios Plato if your first hint can be proved like this:

such that is Setting then set and for all we'd have

Now we get

(add*a*to all)

and the left part of this extended inequality gives us

I guess this is ok, Plato can you respond. And am trying to get the second part. Still, I don't see were this leads. Thx in advance. - September 20th 2010, 02:59 AMMathoMan
Continuing on my previous post. If my getting to is ok then:

(now add )

(compare the reciprocals)

So we get the Now what?

**EDIT:**see my next post, I almost got it ;) - September 20th 2010, 03:17 AMMathoMan
I thought I should put it all in one place.

such that is Setting then set and for all we'd have

Now we get

(add*a*to all)

and the left part of this extended inequality gives us

Then,

(now add )

(compare the reciprocals)

So we get the

If we were to multiply that with

finally I see why comes in handy.

Forcing limit on the above should lead towards solution. Well almost. Strict inequality is bothering me here. Anyway Physix117 that started the thread should work out the fine details. I gave my shot at the rough stuff and I think I did just fine. Waiting for feedback. - September 20th 2010, 04:46 AMHallsofIvy
Physix117, one problem with people not showing any work in their first post is that there are often many different ways to solve a problem, of differing difficulty and mathematical sophistication- and we have no way of knowing which is appropriate for you.

Plato chose to show you the most basic- and therefore, most difficult, method, assuming the**least**he could about your knowledge- that if you are asked to find limits, you at least know the definition of "limit". If I were doing a problem like this, I think I would just observe that , for x> 0, is a**continuous**function and so .

Mathoman, "strict inequality" should not bother you. If for all n, it is NOT true that . What is true is that .

Also, is not**always**true nor is it claimed to be. For "sufficiently large n", . That is, "there exist a positive integer N such that if n> N, then ". That follows directly from the definition of "limit", taking . Since , given any there exist N such that if n> N, then . Taking that becomes which is the same as . Adding a to each part, . Of course, it is only the left part you need here. - September 21st 2010, 04:24 PMmagus
Halls of Ivy and Mathoman. I can see now where plato was leading but there's something that bothers me about the rhs. It involves and I can't get it in a form leading to so I was wondering if something different would work.

Starting from

Is this valid and does it help any? I figure that from the original frac{a}{2}<x_n<\frac{3a}{2} I could find a lower bound with a term in the numerator that would make the limit of it equal 0 and then since

We'd know that

Is this at all valid?