# Prove limit of this recursive sequence

• vrbke1007kkr
In summary: Thanks though! In summary, the homework statement states that there exists an N for every \epsilon> 0, such that n>N implies n>s, which in turn implies that the limit of the sequence is sqrt(s+1).
vrbke1007kkr

## Homework Statement

Let s1=1 and for n>=1 let sn+1=sqrt(sn+1)

Prove that the limit of this sequence is 1/2(1+Sqrt(5))

## Homework Equations

Show that there exist an N for every $$\epsilon$$> 0, such that n>N implies

|Sn-1/2(1+Sqrt(5))|< $$\epsilon$$

## The Attempt at a Solution

I can prove that s is a increasing sequence bounded above thus s converges to a real number, but how to manipulate the terms inside the absolutely value takes a little more than conventional wisdom.

At first we realize that: sn+12 = sn+1, and since we are dealing with an increasing sequence, we can find a bound for sn, but can't see how that's useful

Last edited:
If sn has a limit L, then L has to satisfy L=sqrt(L+1), doesn't it?

thanks, seems exactly what I needed. To reach that conclusion, can I take the following steps:

Let lim sn = s

Since lim (An + Bn)= lim An + lim Bn then lim (Sn + 1) = s + 1

Then use an similar argument for the square root to obtain lim sn+1 = Sqrt(s + 1)

and since lim sn = lim sn+1, s = sqrt(s+1)

vrbke1007kkr said:
thanks, seems exactly what I needed. To reach that conclusion, can I take the following steps:

Let lim sn = s

Since lim (An + Bn)= lim An + lim Bn then lim (Sn + 1) = s + 1

Then use an similar argument for the square root to obtain lim sn+1 = Sqrt(s + 1)

and since lim sn = lim sn+1, s = sqrt(s+1)

Sure. Taking the limit of s_n+1=sqrt(sn+1) gives you L=sqrt(L+1).

If this is for a real analysis course you should probably justify that such a limit must satisfy this equation by mentioning the fact that sqrt(x) is continuous. This is the 'similar' argument you are thinking of when you look at the inference "lim (An + Bn)= lim An + lim Bn then lim (Sn + 1) = s + 1".

sqrt(x) and x+1 are continuous.

ZioX said:
If this is for a real analysis course you should probably justify that such a limit must satisfy this equation by mentioning the fact that sqrt(x) is continuous. This is the 'similar' argument you are thinking of when you look at the inference "lim (An + Bn)= lim An + lim Bn then lim (Sn + 1) = s + 1".

sqrt(x) and x+1 are continuous.

Thanks :) good thing we already proved the latter in a previous theorem so we can use it freely

Equivalently, show that your limit is the root of the equation x^2 - x -1.

Last edited:
ZioX said:
If this is for a real analysis course you should probably justify that such a limit must satisfy this equation by mentioning the fact that sqrt(x) is continuous. This is the 'similar' argument you are thinking of when you look at the inference "lim (An + Bn)= lim An + lim Bn then lim (Sn + 1) = s + 1".

sqrt(x) and x+1 are continuous.

I do not think continuity is need here. These results are simple limit theorems. Besides, continuity usually comes after limit of sequences so I doubt it should be used here.

## 1. What is a recursive sequence?

A recursive sequence is a sequence in which each term is defined by using previous terms in the sequence. This means that each term depends on the term(s) that came before it.

## 2. How do you prove the limit of a recursive sequence?

To prove the limit of a recursive sequence, you need to show that the sequence approaches a specific value as the number of terms increases. This can be done by using mathematical induction or by showing that the terms in the sequence get closer and closer to the limit value as the number of terms increases.

## 3. What is the importance of proving the limit of a recursive sequence?

Proving the limit of a recursive sequence is important because it helps us understand the behavior of the sequence as the number of terms increases. It also allows us to make predictions about the future terms of the sequence and to determine if the sequence is convergent or divergent.

## 4. Is it possible for a recursive sequence to not have a limit?

Yes, it is possible for a recursive sequence to not have a limit. If the terms in the sequence continue to increase or decrease without approaching a specific value, then the sequence is said to be divergent and does not have a limit.

## 5. Are there any special techniques for proving the limit of a recursive sequence?

Yes, there are some special techniques that can be used to prove the limit of a recursive sequence. These include using the squeeze theorem, using the ratio test, and using the root test. It is important to choose the appropriate technique based on the specific sequence being analyzed.

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