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Find the Interval of Convergence

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the interval of convergence.


    2. Relevant equations
    [itex]\displaystyle \sum^{∞}_{n=0} \frac{(x-5)^n}{n^4 * 2^n}[/itex]


    3. The attempt at a solution
    I used the ratio test as follows:

    [itex]\displaystyle \frac{(x-5)^{n+1}}{(n+1)^4 * 2^{n+1}} * \frac{n^4 2^n}{(x-5)^n}[/itex]

    taking the limit:

    (x-5) lim (x->inf) [itex]\displaystyle \frac{n^4}{2(n+1)^4}[/itex] = 0

    Most of these problems the limit comes to 1. How do i solve this now that the limit is zero?
     
  2. jcsd
  3. Apr 4, 2013 #2

    CompuChip

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    You are very much on the right track, except that
    [tex]\lim_{n \to \infty} \frac{n^4}{(n + 1)^4} \neq 0[/tex]

    You can see this if you expand [itex](n + 1)^4 = n^4 + \mathcal O(n^3)[/itex]

    (Don't forget the factor of 1/2 in the end)
     
  4. Apr 4, 2013 #3
    |x-5| < 1

    -1 < x-5 < 1
    4 < x < 6

    Answer (interval of convergence): 4 < x < 6

    Does that look right?
     
  5. Apr 4, 2013 #4
    i think i forgot the factor of (1/2), right?
     
  6. Apr 4, 2013 #5

    HallsofIvy

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    Yes, you did. The correct limit is 1/2. Now, what does that tell you about the radius of convergence?
     
  7. Apr 4, 2013 #6
    what is the radius of convergence vs interval of convergence?
     
  8. Apr 4, 2013 #7
    is this part correct? because that is one of the choices as the answer but 2 < x < 3 is not.
     
  9. Apr 4, 2013 #8
  10. Apr 4, 2013 #9

    CompuChip

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    The radius of convergence is the "width" of the interval of convergence. E.g. if you are looking at the series around x = 2, if the radius of convergence is 4 the interval of convergence would be 2 - 4 < x < 2 + 4, i.e. -2 < x < 6.

    First try answering Halls' question. The radius of convergence is related to the ratio (1/2) you calculated.

    Please don't do that, we are not online 24 hours per day and we will get back to you when we have time.
     
  11. Apr 4, 2013 #10
    sorry for the bump. i'm getting 3 < x < 7 now that I added in the factor of 1/2

    -1 < (x-5)/2 < 1
     
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