Limit of recurrently given sequence

  • Thread starter twoflower
  • Start date
  • #1
368
0
Hi all,

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

Eg. I have this sequence:

[tex]
a_{n} = \sqrt{2}
[/tex]

[tex]
a_{n+1} = \sqrt{2 + a_{n}}
[/tex]

[tex]
\lim_{n \rightarrow \infty} a_{n} = ?
[/tex]

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:

[tex]
\lim_{n \rightarrow \infty} a_{n} = A
[/tex]

[tex]
A = \sqrt{2 + A}
[/tex]

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.
 

Answers and Replies

  • #2
1,357
0
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.
 
  • #3
22
0
I think it is like this:
Since [tex]
\lim_{n \rightarrow \infty} a_{n} = A
[/tex] we can say that [tex]
\lim_{n \rightarrow \infty} a_{n+1} = A_{2}
[/tex].
This gives [tex]
A_{2}=\sqrt{(2+A)}
[/tex] and it´s obvious that [tex]
\lim_{n\rightarrow \infty} a_{n} = \lim_{n\rightarrow \infty} a_{n+1}
[/tex] and then we get [tex]
A_{2}=A=\sqrt{2+A}
[/tex] and then we get A=2.
 
  • #4
1,357
0
Yes. I think that's it! Very good.
 

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