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Convergence of a sequence.

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that [itex] \sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}} [/itex]
    converges and find the limit.
    3. The attempt at a solution
    I can write it also like this correct
    [itex] 2^{\frac{1}{2}},2^{\frac{1}{2}}2^{\frac{1}{4}},2^{\frac{1}{2}}2^{\frac{1}{4}}2^{\frac{1}{8}} [/itex]
    so each time i multiply it by the new number it is getting closer to 1.
    Every new number on the end of the sequence is getting closer to 1 that I am multiplying it by. so this sequence is bounded and decreasing therefore it must have a limit.
    Im not sure what it converges to. but when I know I need to show that it that the sequence
    A-b<ε , where A is the sequence and b is the limit.
     
  2. jcsd
  3. Feb 18, 2012 #2

    Dick

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    No, it's not decreasing. Think about it again.
     
  4. Feb 18, 2012 #3
    I think it does converge. What happens when you multiply exponentials with a common base?

    Edit: this sequence doesn't converge to 1, but...?
     
    Last edited: Feb 18, 2012
  5. Feb 18, 2012 #4
    okay its increasing but it will eventually converge because we keep multiplying it by something smaller. okay I see it now we have 2 raised to the sum of
    [itex] \sum \frac{1}{2^n} [/itex]
    so it should converge to 2 .
    all the exponents should add up to 1.
    by a geometric series.
     
  6. Feb 19, 2012 #5
    You got it! One nitpick though:

    Consider the sequence $$2^{-1}+2, (2^{-2}+2)(2^{-1}+2), (2^{-3}+2)(2^{-2}+2)(2^{-1}+2), \ldots$$
     
  7. Feb 19, 2012 #6

    Dick

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    That's one way to do it. Another way is the call the limit [itex]L[/itex] and figure out what [itex]\sqrt{ 2 L }[/itex] must be if it converges.
     
  8. Feb 19, 2012 #7
    So now I need to show that the sequence [itex] A-2< \epsilon [/itex]
    where A is the sequence. Do I need to write the sequence as 2 raised to a sum with a variable n and then use that to show that it converges. Your saying call the limit L and then put it in and then manipulate it to show that it converges, im mot exactly sure how that would work.
    @alanlu: Are you using that other sequence as an example and I should figure out what it converges to.
    thanks for all the help by the way.
     
    Last edited: Feb 19, 2012
  9. Feb 19, 2012 #8
    The other sequence is an example of a series where the terms get multiplied by decreasing positive numbers, but the series diverges.

    Further hint for the other route: can you write ##\sqrt{2L}## in terms of L?

    Lastly, the definition of the limit of a sequence is a bit unwieldy for this situation: at best, you should use it to check your work. Also, it is stated, sn -> L iff for all e > 0, there is an N such that for all n >= N, ¦L - sn¦ < e.
     
  10. Feb 23, 2012 #9
    Okay now I see what you mean by write in terms of L, write it recursively.
    [itex] x_{n+1}=\sqrt{2x_n} [/itex]
     
    Last edited: Feb 23, 2012
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