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Math Help - Power series convergence

  1. #1
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    Power series convergence

    Hi there. Well, I was trying to determine the radius and interval of convergence for this power series:
    \displaystyle\sum_{0}^{\infty} \displaystyle\frac{x^n}{n-2}

    So this is what I did till now:
    \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{n-1}{n-2}}\right |}=1

    Then \left |{x}\right |<1
    The thing is this result is clearly wrong, I think the series diverges for any x, there is a discontinuity at n=2 for the sum. But I don't know what I'm doing wrong.

    Bye there, thanks for your help!
    Last edited by Ulysses; April 29th 2011 at 07:12 AM. Reason: Changed math to tex delimiters.
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  2. #2
    Ted
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    \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{n-1}{n-2}}\right |}=1
    Where is x here?

    And the sum should NOT start at 0
    It should start at 3 or higher.
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  3. #3
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    It does start at zero. x is not needed to get the radius of convergence. It follows from the ratio test for the convergence of series. What I'm trying to find is the radius for the convergence of the power series, and its interval

    If exists \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}(x-x_0)^{n+1}}{a_{n}(x-x_0)^{n}}}\right |}<1

    Then
    R=\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n}}{a_{n+1}}}\right |}<(x-x_0)

    The thing is I think it actually doesn't exists the limit I gave before. I mean, the series doesn't converges for any x. But I have to demonstrate it.
    Last edited by Ulysses; April 29th 2011 at 07:52 AM. Reason: Corrections, thanks Ted.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ulysses View Post
    It does start at zero.
    It can't start at n = 0. The n = 2 term is singular.

    -Dan
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  5. #5
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    We can take care to say n\ne 2.
    Example: f(x) = \sum\limits_{n = 0}^1 {\frac{{x^n }}{{n - 2}}}  + \sum\limits_{n = 3}^\infty  {\frac{{x^n }}{{n - 2}}}
    Now f(x) is defined for all |x|<1.
    The root test works nicely here.
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  6. #6
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    Ok. I know it diverges at n=2, but in the exercise it starts at n=0. What I'm supposed to do?
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  7. #7
    Ted
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    There is nothing called RATIO FOR THE CONVERGENCE OF THE POWER SERIES !
    Do you mean radius of the convergence or interval of the convergence?

    Your series takes the form \sum \, a_n

    First you will compute \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|.

    In your problem,

    \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^n \cdot (n-2)}{n-1} \right| \rightarrow |x| as n \rightarrow \infty. (Note that we are dealing with x as a constant while evaluating the limit.).

    For the convergence, |x| must be < 1 ,that is -1<x<1

    so we reach the form a<x<b , then the radius equal (b-a)/2 = (1-(-1))/2=2/2=1

    your radius is 1.

    It is impossible for the power series to be divergent for all x.

    at least, it must converge for 1 value for x.


    and the sum should not start at n=0. (your a_n is not continuous at n=2).
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    Yes Ted, I'm sorry, my English sucks :P

    Thank you all, I think that made things clear to me.
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ulysses View Post
    Ok. I know it diverges at n=2, but in the exercise it starts at n=0. What I'm supposed to do?
    The convergence test will still work quite nicely if n starts at 3. (The limit is, after all, taken at infinity.) The summation is just not correctly stated is all.

    -Dan

    Edit: Or you can use Plato's summation. I didn't see that before.
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  10. #10
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    Of course, changing any finite number of terms of an infinite sequence change what it converges to but does not change whether or not it converges.
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