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Power Series - Can someone please help me understand them?

  1. Dec 14, 2004 #1
    I don't understand how to get the inteveral of convergence from a problem:

    for example:

    / -(1)^n(x-2)^n/(n*4^n)

    I know you have to use ratio test and it comes out to:

    The textbook says the series converges when |x-2|<4. How did they come to this conclusion? How do you find c? the answer was c=2. The series converges when -2<x<6 and diverges if either x<-2 or x>6. Can someone explain this also? Thanks
  2. jcsd
  3. Dec 14, 2004 #2
    [tex]\sum_{n=1}^\infty \frac{((-1)^n)(x-2)^n}{n4^n}[/tex] converges when a_n < 1 , so |x-2|/4 < 1 => |x-2|< 4.

    |x-2| < 4 means that for -2 < x < 6, the series converges. plug x=-2 into the series & see that the series converges @ x=-2. plug x=6 into the series & it converges there also. so the interval of convergence is [-2, 6]
  4. Dec 14, 2004 #3

    You just listed the facts and didn't explain them explicitly... anyway I understood most of the stuff through doing the problems. But I still don't understand one thing, when you plug in the x how do you find the limit when n is at infinity, I'm getting confused because of n being the power...

    When you plugged in x=2 how do you solve for it?
  5. Dec 14, 2004 #4

    Tom Mattson

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    They came to it by recognizing that under the ratio test a series converges absolutely if the limit as n goes to infinity of |an+1/an| is less than one. So you get:




    When you work the inequality above into the form:


    then c is the center of the series and R is the radius of convergence.

    Use the definition of the absolute value.


    Now that doesnt' tell you the endpoint convergence. You still have to test for that. So you plug x=-2 and x=6 into your power series (one at a time, of course) and you see if the resultant infinite series converge or diverge. That will tell you which inequality to put an "equal" sign under, if any.
  6. Dec 14, 2004 #5
    Thanks. But you didn't answer my main question, how do I solve the series after plugging in the x value (into the series)?
  7. Dec 14, 2004 #6

    Tom Mattson

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    Have you not studied infinite series??

    If you plug in x=-2, you get:

    [tex]\sum_{n=1}^\infty \frac{((-1)^n)(-2-2)^n}{n4^n}[/tex]
    [tex]\sum_{n=1}^\infty \frac{((-1)^n)(-4)^n}{n4^n}[/tex]
    [tex]\sum_{n=1}^\infty \frac{((-1)^n)(-1)^n4^n}{n4^n}[/tex]
    [tex]\sum_{n=1}^\infty \frac{1}{n}[/tex]

    ...which diverges by the integral test.

    I'll leave it to you to plug in and test the series at x=6.
  8. Dec 14, 2004 #7
    doh! :frown: it's (x-2)^n not (x-2). just ignore my stuff...
  9. Dec 15, 2004 #8


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    There isn't necessarily an expression for the result of a sum. People are probably working on a nice expression for
    [tex]\sum_{i=1}^{\infty} \frac{1}{n^3}[/tex]
    right now.

    All this problem is asking for is convergence, so:

    The ratio between the [itex]n^{th}[/itex] and [itex](n+1)^{th}[/itex] terms is
    [tex]\frac{n+1}{n} \frac{(-1)(x-2)}{4}[/tex]
    in the limit that turns into
    and the ratio must be less than one for the series to converge, so
    [tex]\left|\frac{(-1)(x-2)}{4}\right| < 1[/tex]
    [tex]\left|(x-2)\right| < 4[/tex]
    [tex]-2 < x < 6[/tex]
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