Démontrer la Croissance de (Un+1) / Un

[Sabine]

please help me with this

i'll write it in french bcz i don't know the scientific terms in eng

on considere la suite (Un) definie par:
U1= 1
Un+1= racine de (1 + Un) pour tout entier non nul

Demontrer que (Un+1)\Un > 1

 

[arildno]

If you have:
u_{n+1}=\sqrt{1+u_{n}},u_{1}=1
Then, for ALL n, u_{n}>0 (Agreed?)
Thus, we have: \frac{u_{n+1}}{u_{n}}=\sqrt{\frac{1}{u_{n}^{2}}+1}
What does this tell you?

 

[Sabine]

i agree on un>0 but how did u get the 2nd result?

 

[Gokul43201]

u_{n+1}=\sqrt{1+u_{n}}

Divide both sides by u_n

 

[Sabine]

ok but the second one isn't 1 it's 1\Un

 

[Gokul43201]

That's correct. I'm not sure howArildno got that.

 

[arildno]

AARGH! MY flaw has been discovered before I got to apologize.
Just forget it, Sabine.
Sorry..

 

[Sabine]

it's ok anyway thanks for trying to help

 

[Gokul43201]

Where does the inequality x^2 < x+1 [/itex] hold ? (ie : for what values of x)

 

[Sabine]

0<x<2 but this is not the problem

 

[Sabine]

when x is = to Un

 

[Sabine]

soory i didnt get it arildno

 

[Sabine]

i did not understand i don't want it by reccurency i think this is how u got the result then how did u consider that bn >0

 

[arildno]

There were LATEX errors in my first post; they have now been fixed.

I'm terribly soory; I don't know where mmy mind is today; all I've written is just nonsense.
I'll keep out of this.

 

[HallsofIvy]

Should be easy to do by induction.

u₁= 1 and u₂= \sqrt{2} so u₂> u₁.

Assume u_k+1> u_k for some k. Then u_k+2= \sqrt{1+ u_{k+1}}&gt; \sqrt{1+u_k}= \u_{k+1}.

 

[Sabine]

yes i know but it's not how i want it

 

[A_I_]

hi sabine; je pourrai t'aider.. si tu le veux en francais..
je crois que tu es une etudiante en classe SV?

alors tu as U1= 1
tu calculeras U2 = racine(2)

on suppose que Un est plus grand que 1
alors Un+1 est plus grand que 2
et racine(Un+1) est plus grand que racine(2)

donc on peut diviser (Un+1)/Un et on aura que ce quotient est plus grand que racine(2) d'ou logiquement est plus grand que un


j'espere que tu as compris

 

[Sabine]

non ce n'est pas par recurrence que je veux je le veux sans le raisonnement par recurence

 

[Gokul43201]

0<x<2 but this is not the problem

In fact, the solution is right here.

For U(n+1) to be less than U(n), we must have U(n) > 1.618..., (or negative, which is not allowed) but clearly, the described sequence converges to this number and can hence never exceed it.

 

[A_I_]

pour prouver que U(n+1)/Un plus grand que un

On doit prouver que U(n+1) plus grand que Un

on remarque que U1 plus grand que U0..

Ca doit se faire par recurrence :S?

 

[Sabine]

ouais c par recurence and gokul what's the conclusion?

 

[Gokul43201]

Given :
u_{n+1}=\sqrt{1+u_{n}},u_{1}=1

To prove, for all n,
\frac{u_{n+1}}{u_{n}} &gt; 1

What I had in mind before was not rigorous.

Simply ask youself what the given sequence converges to. As always, the limiting value of the sequence is determined by setting u_{n+1} = u_n, which gives \lim _ {n \rightarrow \infty} u_n = 1.618...

The fact that there is only one positive solution in real n (ie.:1.618...) to the above condition tells you that the sequence is monotonic. Looking at values for n=1,2 tells you that it is monotonically increasing.