Prove limit of this recursive sequence

[vrbke1007kkr]

Homework Statement

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

|S_n-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: s_n+1² = s_n+1, and since we are dealing with an increasing sequence, we can find a bound for s_n, but can't see how that's useful

 

[Dick]

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

 

[vrbke1007kkr]

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

Let lim s_n = s

Since lim (A_n + B_n)= lim A_n + lim B_n then lim (S_n + 1) = s + 1

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

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

 

[Dick]

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

Let lim s_n = s

Since lim (A_n + B_n)= lim A_n + lim B_n then lim (S_n + 1) = s + 1

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

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

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

 

[ZioX]

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.

 

[vrbke1007kkr]

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

 

[╔(σ_σ)╝]

Try using the quadratic equation to get your limit. :-)

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

 

[╔(σ_σ)╝]

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.