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Cauchy sequence

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data

    Show that the following sequence converges:

    xn= (sin(1)/2) + (sin(2)/2^2) + (sin(3)/2^3) +...+ (sin(n)/2^n)
    2. Relevant equations



    3. The attempt at a solution
    To show that it converges, i want to show that it is a cauchy sequence (since all cauchy sequences converge).
    I know that xn is cauchy if abs(xn-xm)< E for all E>0.
    and the above sequence can be written as:
    [tex]\sum(sink/2^k[/tex]

    But i dont know how to proceed??
    Any help would be very much appreciated.
     
  2. jcsd
  3. Oct 25, 2008 #2
    Is that a fact? Check really if that's true...

    As for your sequence, your sequence is (monotonically) increasing and bounded thus convergent. Can you show that it is true?
     
  4. Oct 25, 2008 #3

    Dick

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    It's not monotonic. sin(k) can be negative. And why are you questioning whether cauchy sequences converge?? The question is clearly about the real numbers, and they are complete. All cauchy sequences DO converge. sara_87, to show it's cauchy just notice sin(k)/2^k<=1/2^k. You can bound abs(xn-xm) by the sum of a geometric series.
     
  5. Oct 25, 2008 #4
    Yes, you're right I made two mistakes.

    Dick uses [tex] sin(x) \leq 1[/tex]
     
  6. Oct 25, 2008 #5

    Dick

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    Yep. And I should have written |sin(k)/2^k|<=1/2^k.
     
  7. Oct 25, 2008 #6
    ok, so |sin(k)/2^k|<=1/2^k

    and the formula for geometric series is:
    a(1-r^n)/1-r

    i dont know how to apply this to sin(k)/2^k
     
  8. Oct 25, 2008 #7
    Look, you want to have [tex]|a_n - a_m|< \varepsilon [/tex].

    Now you know you can make an upper bound via a geometric series, okay?
     
  9. Oct 25, 2008 #8
    Yes, i understand this bit and i know that in this case, we have:
    abs(sin(k)/2^k) < 1/2^k

    but is this it?? surely not. I still havent proved that it is cauchy.
     
  10. Oct 25, 2008 #9

    Mark44

    Staff: Mentor

    The problem as stated asked that you prove that the sequence was convergent. You brought up the part about Cauchy sequences. If you can show that the sequence converges (see Dick's and Dirk's posts), you're done.
     
  11. Oct 25, 2008 #10
    Yes, but i 'want' to show that it is cauchy (part of the question). I know that:
    abs(xn-xm)< E
    and
    abs(sin(k)/2^k) < 1/2^k
    but how do i relate these two?
     
  12. Oct 25, 2008 #11

    Mark44

    Staff: Mentor

    Using your definition for xn at the beginning of this thread, what is xn - xm?
     
  13. Oct 25, 2008 #12
    (sin(1)/2) + (sin(2)/2^2) + (sin(3)/2^3) +...+ (sin(n)/2^n) - xm
     
  14. Oct 25, 2008 #13

    Dick

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    Let's assume n>m. So xn-xm=sin(m+1)/2^(m+1)+sin(m+2)/2^(m+2)+...sin(n)/2^n. |xn-xm|<=1/2^(m+1)+...+1/2^n. Apply your geometric series thing to that.
     
  15. Nov 1, 2008 #14
    ok, so i get:
    abs(xn-xm)<=1/2^m for all m>N ;
    but this must mean that 1/2^m is less than E but how do we know that m is greater than E (for this to be true)?
     
  16. Nov 1, 2008 #15

    Dick

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    You are given e. You PICK an N large enough that 1/2^N<e. That means |xn-xm|<e for all n,m>N, right? That's Cauchy.
     
  17. Nov 3, 2008 #16
    but however large N is, the 1/2^N will always be postive.
    so, just as an example, what kind of value can N be?
     
  18. Nov 3, 2008 #17
    N will depend on your epsilon chosen. In your case you should use the geometric series to find your N.
     
  19. Nov 3, 2008 #18

    Dick

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    Pick N so 2^N>1/epsilon.
     
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