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Convergence series interval

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data

    First of all, in the below equation, it wont let me display it correctly. In the denominator is supposed to display 3^(n-1), NOT 3(n-1).

    So, I need to find the convergence interval and check the points. So far, I've found the intervals and checked both of the points to converge to 0. So my interval I obtained is [-13/3, -11/3].

    I would love input as to whether or not I'm correct.


    2. Relevant equations

    [tex]\sum[/tex] [tex]\frac{(x+4)^n}{(2n-1)3^(n-1)}[/tex]

    (Remember, it's supposed to read (...)3^(n-1)

    3. The attempt at a solution

    I got to (6n-3)/(2n+1) = 6/2 = 3, so.... =3|x+4| < 1

    |x+4| < 1/3 (series converges) ....

    -1/3 < x+4 < 1/3

    -13/3 < x < -11/3
     
  2. jcsd
  3. Apr 21, 2010 #2

    Dick

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    I think you have something upside down in your ratio test. I get |x+4|/3<1. And I also don't get that both endpoints converge. Can you try that again and show your work?
     
  4. Apr 21, 2010 #3
    Sure thing! Here's everything I've got....


    [tex]\sum[/tex] [tex]\frac{(x+4)^n}{(2n-1)3^(n-1)}[/tex]

    (again, it's supposed to be (...)3^(n-1) )

    Now I do (or try to) do the ratio test....
    L = LIM [tex]\frac{(x+4)^n+1}{(2n+1)3^(n)}[/tex][tex]\frac{(2n-1)3^(n-1)}{(x+4)^n}[/tex]

    From that (and it isn't displaying properly), I do some canceling, and end up with....

    L= LIM [tex]\frac{3(x+4)(2n-1)}{(2n+1)}[/tex]

    I then take out the 3(x+4) and put it before the LIM....


    L= 3(x+4) LIM [tex]\frac{2n-1}{2n+1}[/tex]

    Not sure what else to do with that, I use L'Hospital's rule and get 2/2 = 1

    Then....

    = 3|x+4| < 1

    = |x+4| < 1/3

    so...

    -1/3 < x+4 < 1/3

    -13/3 < x < -11/3


    It didn't look quite right to me... any guidance would be hugely appreciated. Thanks!
     
  5. Apr 21, 2010 #4
    I got -7 < x < -1 and my end points don't converge.

    You messed up your algebra,[tex]3^{n-1} / 3^{n} [/tex] is 1/3
     
  6. Apr 21, 2010 #5
    Wow.... yep, needless to say you are most definitely correct... the algebra always gets me! Let me re-work this now....

    Thank you!
     
  7. Apr 22, 2010 #6
    After going through it again, I do indeed get |x+4|/3 < 1, so....

    -3 < x+4 < 3

    finally....

    -7 < x < -1


    Checking the points..... plugging -7 in I do find it converges to 0.

    Plugging -1 in and doing a ratio test again, I get it converges to 1, hence diverges. My final divergent quotient is:

    3(2n-1) / 3(2n+1) = 1

    .... hopefully that is the correct way to do it... I'm learning all kinds of things tonight, some things make more sense than others! ;-) Thanks guys!
     
  8. Apr 22, 2010 #7

    Dick

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    Of course the ratio test is going to give you 1 at the endpoints. But that doesn't mean it diverges there. It's means the ratio test is inconclusive. You have to do another test at the endpoints. What's the form of the series at each endpoint?
     
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