Limit of recurrently given sequence

[twoflower]

Hi all,

I don't fully understand solving of limits when the sequence is given by some recurrent expression.

Eg. I have this sequence:

<br /> a_{n} = \sqrt{2} <br />

<br /> a_{n+1} = \sqrt{2 + a_{n}} <br />

<br /> \lim_{n \rightarrow \infty} a_{n} = ?<br />

First, I should prove the monotony and finiteness (is it ok to say it in english this way?). Well I did the proof the monotony by induction, I hope right. Now the finiteness. How should I do it? Can I just guess it won't get greater than. let's say 2 ? Ok, I chose 2 and prove that it is finite.

Now the limit. Our teacher wrote this:

<br /> \lim_{n \rightarrow \infty} a_{n} = A<br />

<br /> A = \sqrt{2 + A}<br />

And I ask, what should this mean? Where does this equality come from?

Of course to solve it is easy and we find out A = 2, which is the limit. But I don't understand why the equality.

Thank you for any help.

 

[e(ho0n3]

If the limit exists then the difference between a[n+1] and a[n] for very, very large values of n is very, very small. In other words, a[n+1] is approximately a[n] for very large n. Hence the substitution. I don't find this very satisfying though. Maybe somebody else can provide a more rigoruous explanation.

 

[danja347]

I think it is like this:
Since <br /> \lim_{n \rightarrow \infty} a_{n} = A<br /> we can say that <br /> \lim_{n \rightarrow \infty} a_{n+1} = A_{2}<br />.
This gives <br /> A_{2}=\sqrt{(2+A)}<br /> and it´s obvious that <br /> \lim_{n\rightarrow \infty} a_{n} = \lim_{n\rightarrow \infty} a_{n+1}<br /> and then we get <br /> A_{2}=A=\sqrt{2+A}<br /> and then we get A=2.

 

[e(ho0n3]

Yes. I think that's it! Very good.