Baby rudin, Limit of nth root of p

[jecharla]

Part (b) of theorem 3.20 is to prove that the limit as n approaches infinity of the nth root of p equals one(for p>0). The proof given in the text uses some inequality derived from the binomial theorem which seems to me to just come out of nowhere and provide a completely unintuitive proof. That's great that this inequality works but how would I know to use this specific inequality in the first place? Could someone either provide a more explanatory proof which perhaps does not use the binomial theorem, or clarify the proof in the text so that it doesn't seem so unnatural?

 

[STEMucator]

Part (b) of theorem 3.20 is to prove that the limit as n approaches infinity of the nth root of p equals one(for p>0). The proof given in the text uses some inequality derived from the binomial theorem which seems to me to just come out of nowhere and provide a completely unintuitive proof. That's great that this inequality works but how would I know to use this specific inequality in the first place? Could someone either provide a more explanatory proof which perhaps does not use the binomial theorem, or clarify the proof in the text so that it doesn't seem so unnatural?

Recall that the nth root can be re-written like so :

\sqrt[n]{p} = p^1/n

Recall further that the lim n-> ∞ 1/n = 0 so we know that lim n-> ∞ \sqrt[n]{p} = 1 since anything to the power of zero is one.

Something you could do alternatively is use the M, N definition of a limit going to infinity.

That should help.

 

[Integral]

Perhaps you need to spend a little more time staring at that proof.
( 1 + x_n)^n=p is a algebraic restatement of x_n = {^n\sqrt{p}}-1

The RHS of the inequality is simply the first 2 terms of the exansion of ( 1 + x_n)^n . Since he has dropped many postitive terms what is left is smaller then the full expansion, thus the inequality.

The different expansions need to be a tool on your mathematical belt, get used to it and learn to use them.

I observed this fact within a few months of picking up my first scienctic calculator. No matter where I started, repetive square root operations ended up at 1.

 

[Bacle2]

Recall that the nth root can be re-written like so :

\sqrt[n]{p} = p^1/n

Recall further that the lim n-> ∞ 1/n = 0 so we know that lim n-> ∞ \sqrt[n]{p} = 1 since anything to the power of zero is one.

Something you could do alternatively is use the M, N definition of a limit going to infinity.

That should help.

.

This is a good idea, but it needs the assumption of continuity; I don't have Rudin--

nor his book ;) --with me, I don't know if he proved continuity of x^1/n

by this point. I don't see how else you can justify f(lim n→∞)=lim_n→∞

applied to x^1/n

 

[jecharla]

Integral,

I think you are right that I need to spend more time staring at this proof(which I have been doing for quite a while now).

I have no problem understanding this proof in that I can follow the algebra. However I don't understand how one could come up with this proof from scratch. Often in order to truly understand a proof, you have to not just be able to follow the reasoning but also understand how it was thought that this type of reasoning should be applied in the first place. I have stared at this proof for so long that I can easily replicate it and understand why it must be true(because the equation provides the desired result) but if I was faced with a different problem requiring the same technique I would be lost. Is there some sort of "scratch work" which is left out here which would lead one to this proof?

 

[Bacle2]

Integral,

I think you are right that I need to spend more time staring at this proof(which I have been doing for quite a while now).

I have no problem understanding this proof in that I can follow the algebra. However I don't understand how one could come up with this proof from scratch. Often in order to truly understand a proof, you have to not just be able to follow the reasoning but also understand how it was thought that this type of reasoning should be applied in the first place. I have stared at this proof for so long that I can easily replicate it and understand why it must be true(because the equation provides the desired result) but if I was faced with a different problem requiring the same technique I would be lost. Is there some sort of "scratch work" which is left out here which would lead one to this proof?

I don't know if this is what you are looking for, but a more general approach I can think

of is using the fact (check) that the sequence x^1/n (for fixed x, of course) is

monotone and bounded, so that it must converge --to its lub. Re inequalities, I don't

see how to avoid just learning a bunch of them, as using inequalities is the bread-and-

butter of Standard Analysis.

 

[Mute]

I have no problem understanding this proof in that I can follow the algebra. However I don't understand how one could come up with this proof from scratch. Often in order to truly understand a proof, you have to not just be able to follow the reasoning but also understand how it was thought that this type of reasoning should be applied in the first place. I have stared at this proof for so long that I can easily replicate it and understand why it must be true(because the equation provides the desired result) but if I was faced with a different problem requiring the same technique I would be lost. Is there some sort of "scratch work" which is left out here which would lead one to this proof?

Figuring out how to do a proof 'from scratch' comes from experience and having tried lots of different things (that didn't pan out).

 

[Skrew]

This proof is easy, use the inf of the set. The proof involveing n^(1/n) is the one that involves the binomial theorem.