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Does the series converge?

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Find all x for which [tex]\sum[/tex] from k=1 to infinity (x^k - x^(k-1))(x^k+x^(k-1)) converges.


    2. Relevant equations
    I think the geometric series formula is relevant: [tex]\sum[/tex] k=N to infinity of x^k = 1/(1-x) for all |x|<1.


    3. The attempt at a solution
    I simplified the expression to x^2k - x^(2k-2). I can show that the first term converges (I think ... because it is the product of two convergent sequences?), and I understand logically why the second term converges, but not sure how to show rigorously. I also think that the series converges for all |x|<1, but again not sure how to construct the proof.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 15, 2009 #2

    LCKurtz

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    Hint: Factor x(k-1) out.
     
  4. Nov 16, 2009 #3
    But then x^2 is one of the factors, and x^2 does not converge (?).
     
  5. Nov 16, 2009 #4
    You are summing over k, not over x.

    Hint number 2: your guess for the values of x where the series converges is almost right, but not quite.
     
  6. Nov 16, 2009 #5
    Ah, I see, thank you. In this case, unlike the "regular" geometric series case, x can be less than one OR one (i.e., |x|<=1), because if x is one, x^2k and x^(2k-1) are both one ... so the series converges to zero. Correct?
     
  7. Nov 16, 2009 #6

    LCKurtz

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    Yes, it converges at x = 1. But if you follow my hint and factor out x(k-1) in your original expression, there won't be any k left in the summand. What does that tell you?
     
  8. Nov 16, 2009 #7
    OK, I think I understand: the expression factors to x^(k-1)*[x^(k-1)-x^(k-1)] = 0?
     
  9. Nov 16, 2009 #8

    LCKurtz

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    In the summation you started with

    [tex]\frac {x^k - x^{k-1}}{x^k+x^{k-1}}[/tex]

    It's a fraction now, and it will be a fraction after you factor x(k-1) out and simplify it. Do that and simplify it. What do you get?
     
  10. Nov 16, 2009 #9
    There is no fraction in the original summation.
     
  11. Nov 16, 2009 #10

    LCKurtz

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    Woops! Sorry, I misread the problem.
     
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