# Homework Help: Me with this

1. Apr 26, 2005

### Sabine

plz 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

Last edited: Apr 26, 2005
2. Apr 26, 2005

### 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?

3. Apr 26, 2005

### Sabine

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

4. Apr 26, 2005

### Gokul43201

Staff Emeritus
$$u_{n+1}=\sqrt{1+u_{n}}$$

Divide both sides by $u_n$

5. Apr 26, 2005

### Sabine

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

6. Apr 26, 2005

### Gokul43201

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

7. Apr 26, 2005

### arildno

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

8. Apr 26, 2005

### Sabine

it's ok anyway thanks for trying to help

9. Apr 26, 2005

### Gokul43201

Staff Emeritus
Where does the inequality $$x^2 < x+1 [/itex] hold ? (ie : for what values of x) 10. Apr 26, 2005 ### Sabine 0<x<2 but this is not the problem 11. Apr 26, 2005 ### Sabine when x is = to Un 12. Apr 26, 2005 ### Sabine soory i didnt get it arildno 13. Apr 26, 2005 ### Sabine i did not understand i don't want it by reccurency i think this is how u got the result then how did u consider that bn >0 14. Apr 26, 2005 ### arildno There were LATEX errors in my first post; they have now been fixed. I'm terribly soory; I don't know where mmy mind is today; all I've written is just nonsense. I'll keep out of this. Last edited: Apr 26, 2005 15. Apr 26, 2005 ### HallsofIvy Should be easy to do by induction. u1= 1 and u2= [tex]\sqrt{2}$$ so u2> u1.

Assume uk+1> uk for some k. Then uk+2= $$\sqrt{1+ u_{k+1}}> \sqrt{1+u_k}= \u_{k+1}$$.

16. Apr 26, 2005

### Sabine

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

17. Apr 26, 2005

### 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

18. Apr 26, 2005

### Sabine

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

19. Apr 27, 2005

### Gokul43201

Staff Emeritus
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.

20. Apr 28, 2005

### 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?