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Homework Help: Interval of Convergence

  1. Nov 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the interval of convergence of each of the following
    Ʃ[itex]^{∞}_{n=0}[/itex] ([itex]\frac{3^{n}-2^{2}}{2^{2n}}[/itex](x-1)[itex]^{n}[/itex])


    2. Relevant equations
    Please refer to attachment


    3. The attempt at a solution
    Please refer to attachment. All I want to know is that I'm doing this problem right. I have found the interval but haven't plugged the interval back into the equation.
     

    Attached Files:

  2. jcsd
  3. Nov 25, 2011 #2

    Dick

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    You shouldn't break it into two series like that unless you know what you are doing. And combining them into (3/4)(x-1)-(1/4)(x-1) shows you probably don't. Just apply the ratio test to the whole expression.
     
  4. Nov 25, 2011 #3
    The thing is I don't know what to do with that 2^2 in the equation, it throws me off. I'll re-post a picture this time everything together.
     
  5. Nov 25, 2011 #4

    Dick

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    One of the factors in your ratio test should be (3^(n+1)-2^2)/(3^n-2^2), right? To find the limit of that as n->infinity, just divide numerator and denominator by 3^n.
     
  6. Nov 26, 2011 #5
    I tried your suggestion, however I didn't come to a good end (maybe I did something wrong). Please take a look at the attached file; this is my re-attempt.
     

    Attached Files:

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  7. Nov 26, 2011 #6

    HallsofIvy

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    It looks more complicated than necessary- it should be obvious that
    [tex]\frac{3^{n+1}- 4}{3^n- 4}[/tex]
    is dominated by [itex]3^{n+1}{3^n}= 3[/itex] and so its limit is 3. But your result, that the limit is [itex](3/4)|x- 1|[/itex] is correct. Now, what is the answer to your original question?
     
  8. Nov 26, 2011 #7

    Dick

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    Not right. There's a mistake in the long division. 3^(n+1)-3^n isn't 3. And I didn't mean that in my suggestion. Take (3^(n+1)-4)/(3^n-4). Dividing the numerator, 3^(n+1)-4 by 3^n gives you 3^(n+1)/3^n-4/3^n. What's 3^(n+1)/3^n? Now take the limit as n->infinity. Do the same with the denominator.
     
  9. Nov 28, 2011 #8
    Thank you I got it now.

    Thanks
     
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