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Limit of recurrently given sequence

  1. Nov 29, 2004 #1
    Hi all,

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

    Eg. I have this sequence:

    a_{n} = \sqrt{2}

    a_{n+1} = \sqrt{2 + a_{n}}

    \lim_{n \rightarrow \infty} a_{n} = ?

    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:

    \lim_{n \rightarrow \infty} a_{n} = A

    A = \sqrt{2 + A}

    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.
  2. jcsd
  3. Nov 29, 2004 #2
    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.
  4. Nov 29, 2004 #3
    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}
    This gives [tex]
    [/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]
    [/tex] and then we get A=2.
  5. Nov 29, 2004 #4
    Yes. I think that's it! Very good.
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