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Real Analysis proof help, convergence

by CrazyCalcGirl
Tags: analysis, convergence, proof, real
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CrazyCalcGirl
#1
Sep21-07, 11:51 AM
P: 15
1. The problem statement, all variables and given/known data

The problems states let S1 = 2 and S(n+1) = (square root(2Sn+1)) show that Sn is convergent.

2. Relevant equations



3. The attempt at a solution

I am not really sure how to go about this proof. Sometimes it is easier to prove something is cauchy than to show it is convergent, but cauchy sequences are a very new topic for me so I need some guidance. If I do need to show its cauchy just simply because that is much easier than showing regular convergence what would the proof look like? My RA text book is awful and gives no real examples.
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Dick
#2
Sep21-07, 11:59 AM
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It's often handy to get out a calculator and punch out the first few terms in the sequence. It looks to me like the sequence is increasing and has an upper bound. It's also handy to notice that if L is the limit, then L probably satisfies L=sqrt(2L+1). Now can you prove these statements?
CrazyCalcGirl
#3
Sep21-07, 12:26 PM
P: 15
ok.. So I can see by plugging it in on my calculator that it is definitely increasing and I'm getting the limit to be about 2.41421. Is that what you got? Ok so if I can show it is bounded above (which it obviously is) and it is a monotone increasing sequence then it converges by the monotone convergence theorem. I guess that is not as bad as I thought. He just gave us this homework after he taught cauchy sequences so I figured he wanted us to try to use that.

Dick
#4
Sep21-07, 01:04 PM
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Real Analysis proof help, convergence

Right. Now you just have to do some algebra to show it's increasing and bounded.
CrazyCalcGirl
#5
Sep21-07, 01:33 PM
P: 15
I've got the increasing part down. I started off with S(n+2)/S(n+1) >1 and basically used algebra to get it down to S(n+1) > Sn. That is where I stopped. I'm think that is enough to show its increasing for all n >=x. Now for the bounded above part. I am not sure this part works.

I knew it was bounded above by 2.41 so can I use induction to show Sn has an upper bound for all terms? Say I arbitrarily pick 3 to start with.

So I know
1. S1 <=3 because 2 <=3.

2. Assume Sn<=3

3. Show S(n+1) <=3. I end up with S(n+1)<= (Sq root(7))<=3 since that is true it must be true for all n.

I am not having trouble finishing the induction. I just do not know if this works to show something is bounded above. Our book doesn't show any examples on this. It just gives theorems and then homework. :P
Dick
#6
Sep21-07, 01:46 PM
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I think you are making this more complicated than it is. To show the sequence is increasing you need to show sqrt(2x+1)>x. What range of x is this true for? To show bounded above then find the true upper bound (and limit!) L, and show if x<L then sqrt(2x+1)<L.
CrazyCalcGirl
#7
Sep21-07, 02:55 PM
P: 15
hmm Well the problem I'm having is that my professor will not allow us to use anything at all if we have not done it in analysis yet. I used induction because it was in a prior section. I'm not sure I've seen an example of how to find the true upper bound.

We've done problems where we were given a limit and asked to prove it using the definition of convergence, but I've not seen one yet on how to find the upper bound. If you gave me a sequence I could find the supremum just by plugging in terms and seeing where it goes. That's how I got the 2.14, but without using numbers I'm stuck.
Dick
#8
Sep21-07, 02:58 PM
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As I said, a limit L is a fixed point of the sequence, so you should have L=sqrt(2L+1). What's L? This problem is less about deep principles and more about algebraic cleverness.
CrazyCalcGirl
#9
Sep21-07, 05:45 PM
P: 15
ok well .. If you want me to solve for L I moved it over and completed the square..then I get L=+/- (sq root(2)) +1 so the positive number matches the 2.14 I found before. So when you asked me to show sqrt(2x+1)>x you get the same thing with an inequality sign. The range that this is true for is when x is>=(-1/2).

I still don't think I'm getting what to do with show "sqrt(2x+1)>x"

I'm trying to solve that in a meaningful way, but I just get it down to +/-(sq root(2) + 1) > x. Maybe solving for x isnt the point in this.
Dick
#10
Sep21-07, 05:55 PM
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The range in which sqrt(2x+1)>x is (-1/2,L) where L=1+sqrt(2). Doesn't this say that your sequence is increasing as long as it's values stay in that range? Your starting point of 2 is in that range. Now you just have to show that the sequence never leaves that range. Now you just want to show that if x is in (-1/2,L) then sqrt(2x+1) is also in (-1/2,L).
rikomoja
#11
Sep17-09, 06:05 PM
P: 1
Please help
suppose k>3, x, y R^k, |x − y| =d>0,and r>0. Prove

If 2r>d, there are infinitely many z R^k such that

|z-x| =|z − y| = r
Dick
#12
Sep17-09, 10:14 PM
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Quote Quote by rikomoja View Post
Please help
suppose k>3, x, y R^k, |x − y| =d>0,and r>0. Prove

If 2r>d, there are infinitely many z R^k such that

|z-x| =|z − y| = r
This has nothing to do with the original thread. If you have question, start your own thread. In that new thread try to contribute some ideas about why that might or might not be true.


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