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Math Help - Interesting technique to find a limit...

  1. #1
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    Interesting technique to find a limit...

    Suppose I wanted to find the limit of some sequence, given by a_{n+1} = \sqrt{2a_n}. Suppose also that for this particular sequence, we have a_n \leq a_{n+1} < 2; i.e., the sequence in increasing and converges to 2. Rearranging the sequence would give \frac{a_{n+1}^2}{a_n} = 2. But \displaystyle \lim _{n \to \infty} a_n = \lim _{n \to \infty} a_{n+1} = L, and so \displaystyle \lim _{n \to \infty} (a_n) = \lim _{n \to \infty} \frac{a_{n+1}^2}{a_n} = \frac{L^2}{L} = L = 2.

    I already know that this is the correct answer, but is there anything wrong with my methodology?
    Last edited by CaptainBlack; August 24th 2010 at 02:27 AM.
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  2. #2
    Newbie
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    Jan 2010
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    No. Note, however, two things:

    1. You missed a corner case allowing L = 0 as an additional solution.
    2. You've only proven that *if* the sequence converges, then it must converge to 2 or 0. You still need to show that it does indeed converge (in this case for any initial value a_0 >= 0). In this case that's easy, thanks to some useful inequalities satisfied by the geometric mean.

    For example, a similar argument shows that the sequence a_n = -a_{n-1} + 1 formally converges to 1/2, but in fact that's only true for a_0 = 1/2. The sequence fails to converge for all other initial values.
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