Convergent Sequence of Square Roots

gajohnson
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Homework Statement



Let S_{1}=1 and S_{n+1}=\sqrt{2+S_n}

Show that \left\{S_n\right\} converges and find its limit.

Hint: First assume that the limit exists, then what is the possible value of the limit? Second, show that the sequence is increasing and bounded. Finally, follow the definition of convergence to show that the sequence converges.

Homework Equations



NA

The Attempt at a Solution



Well it is pretty clear that this converges to 2, so that's a start.

I am having difficulty constructing a good way to show that the sequence is increasing and bounded. Any help getting started would be nice.

Thanks!
 
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gajohnson said:

Homework Statement



Let S_{1}=1 and S_{n+1}=\sqrt{2+S_n}

Show that \left\{S_n\right\} converges and find its limit.

Hint: First assume that the limit exists, then what is the possible value of the limit? Second, show that the sequence is increasing and bounded. Finally, follow the definition of convergence to show that the sequence converges.

Homework Equations



NA

The Attempt at a Solution



Well it is pretty clear that this converges to 2, so that's a start.

I am having difficulty constructing a good way to show that the sequence is increasing and bounded. Any help getting started would be nice.

Thanks!

To show it's increasing you want to show sqrt(x+2)>x, right? For what range of x is that true? Try to solve the inequality.
 
Dick said:
To show it's increasing you want to show sqrt(x+2)>x, right? For what range of x is that true? Try to solve the inequality.

Well because S_1=1 is given, the sequence is strictly increasing for x\in[1,2), and the sequence is monotonically increasing for x\in[1,2].

Is showing this by solving the inequality enough to claim that the sequence is increasing and also bounded by 2 (since solving the above as an equality gives 2)?
 
gajohnson said:
Well because S_1=1 is given, the sequence is strictly increasing for x\in[1,2), and the sequence is monotonically increasing for x\in[1,2].

Is showing this by solving the inequality enough to claim that the sequence is increasing and also bounded by 2 (since solving the above as an equality gives 2)?

Yes, showing the inequality for the range x in [1,2) will show it. To show it's bounded you need to show the inequality sqrt(x+2)<2 holds in that range.
 
Dick said:
Yes, showing the inequality for the range x in [1,2) will show it. To show it's bounded you need to show the inequality sqrt(x+2)<2 holds in that range.

I believe I've got it now. Thanks for your help!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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