# Démontrer la Croissance de (Un+1) / Un

• Sabine
In summary: However, if you were to ask for the limit of the sequence as n gets larger, you would get an infinite value, since there could be an infinite number of solutions that are larger than 1.618...
Sabine

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:
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?

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

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

Divide both sides by $u_n$

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

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

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

it's ok anyway thanks for trying to help

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

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

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

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

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

Sabine said:
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.

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?

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

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

To prove, for all n,
$$\frac{u_{n+1}}{u_{n}} > 1$$

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.

## 1. What is the formula for demonstrating the growth of (Un+1) / Un?

The formula for demonstrating the growth of (Un+1) / Un is (Un+1) / Un = (1 + r) where r is the growth rate.

## 2. How is the growth rate (r) calculated in (Un+1) / Un?

The growth rate (r) in (Un+1) / Un is calculated by taking the difference between the final value (Un+1) and the initial value Un, and then dividing it by the initial value Un. It can also be calculated by taking the natural logarithm of (Un+1) / Un.

## 3. What does a positive growth rate (r) indicate in (Un+1) / Un?

A positive growth rate (r) in (Un+1) / Un indicates that the value of (Un+1) is greater than the value of Un, and there is an overall increase in the sequence.

## 4. How does the growth rate (r) affect the overall growth of (Un+1) / Un?

The growth rate (r) has a direct impact on the overall growth of (Un+1) / Un. A higher growth rate indicates a faster rate of growth, while a lower growth rate indicates a slower rate of growth.

## 5. What is the significance of demonstrating the growth of (Un+1) / Un?

Demonstrating the growth of (Un+1) / Un is important in understanding the trend and pattern of a sequence. It can also help in making predictions about future values and analyzing the performance of a particular system or process.

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