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Homework Help: Power series

  1. Nov 28, 2004 #1
    For the most part I seem to understand what I'm doing, but whenever you toss in a [tex](-1)^n[/tex] or a [tex](-1)^{n+1} [/tex] it starts to trip me up.
    For example:

    [tex]\sum\frac{(-1)^{n+1}(x-5)^n}{n5^n}[/tex]

    pretty straight forward, it's centered at 5, use the ratio test and solve the inequality to obtain R = 5 with endpoints 0, 10. It's when I go to test the endpoints that I get into trouble. I am going to solve this the way I would and maybe someone can point out where I'm going wrong with my thinking or algebra.

    [tex]\sum\frac{(-1)^{n+1}(0-5)^n}{n5^n}[/tex]

    [tex]\sum\frac{(-1)^{n+1}(-5)^n}{n5^n}[/tex]

    Just as I was typing this, I think I found my mistake, but how do I fix it?
    I have been factoring out a [tex]-1[/tex] from the [tex](-5)^n[/tex] and canceling the [tex]5^n[/tex] with the one in the denominator. You can't do that can you? How about this?

    [tex]\sum\frac{(-1)^{n+1}(-1)^n(5)^n}{n5^n}[/tex]

    That would let the 5^n's cancel and would leave

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


    right? Isn't this still an alternating series? The solutions manual says that when x=0 it should go to

    [tex]\sum\frac{-1}{n}[/tex]

    which is a divergent p series and when x = 10 it stays an alternating series.
     
  2. jcsd
  3. Nov 28, 2004 #2
    You are right. When x=0 you get a divergent p series. But the [tex]\sum\frac{(-1)^{2n+1}}{n}[/tex] is NO alternating series because the sign always stays minus. If x = 10, you will get an alternating series : [tex]\sum\frac{(-1)^{n+1}}{n}[/tex]

    regards
    marlon
     
  4. Nov 28, 2004 #3
    Thanks marlon, I see it now. No matter what n is, your always going to get -1 raised to an odd value.
     
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