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Power Series

  1. Jan 31, 2009 #1
    Find the interval of convergence of the given series and its behavior at the endpoints:

    [tex]\sum^{+\infty}_{n=1} \frac{(x+1)^n}{\sqrt{n}}[/tex] [tex]= (x+1) + \frac{(x+1)^2}{\sqrt(2)}+...[/tex]

    The attempt at a solution

    Using the ratio test: [tex]\left|\frac{S_{n+1}}{S_{n}}\right| = \sqrt{\frac{n}{n+1}}\left|x+1\right|[/tex]

    Hence, [tex]lim_{n\rightarrow\infty} \left|\frac{S_{n+1}}{S_{n}}\right|= \left|x+1\right|[/tex]

    So, the interval of convergence is |x+1|<1, which implies -1<x<1 which in turn is equal to: -2<x<0

    At the right hand endpoint where x = 0 we have the divergent p-series [tex]\sum^{+\infty}_{n=1} \frac{1}{\sqrt{n}}[/tex] (with p=1/2).

    At the left endpoint x=-2 we get the alternating series [tex]\sum^{+\infty}_{n=1} \frac{(-1)^n}{\sqrt{n}}[/tex]. The book says this series converges but I used the comparison test & found out that it's NOT!

    Sn = (1)n+11/√n diverges by comparison with the p-series ∑1/√n (since p<1)

    I’m really confused right now, if it is supposed to converge then why does it diverge by the comparison test? What mistakes did I make?

  2. jcsd
  3. Feb 1, 2009 #2
    I'm confused by what you are comparing [tex] \frac{(-1)^n}{\sqrt{n}} [/tex] to?

    How can you compare it to [tex] \frac{1}{\sqrt{n}} [/tex]?

    The first has negative and positive terms whereas the latter does not.
  4. Feb 1, 2009 #3
    So, what else can I compare it to.....?

    Why can't we just re-write the first as [tex](-1)^{n+1}\frac{1}{\sqrt{n}}[/tex] and then compare it to the p-series [tex]\sum \frac{1}{\sqrt{n}}[/tex]?
  5. Feb 1, 2009 #4
    because [tex] (-1)^{n+1} \frac{1}{\sqrt{n}} [/tex] will still have positive and negative terms right? You'd need a test that has "Absolute" in the name :) if anything.
  6. Feb 1, 2009 #5
    Hmm, I'm really curious!...
    For example, what test can you use for [tex]\frac{(-1)^n}{\sqrt{n}}[/tex]?

  7. Feb 1, 2009 #6


    Staff: Mentor

    How about the alternating series test? You know that one, don't you?
  8. Feb 1, 2009 #7
    Oh OK, here's a part of the theorem for alternating series: Let ∑(-1)n+1an be an alternating series. If the sequance [tex]\left\langle a_{n}\right\rangle[/tex] is decreasing then ∑(-1)n+1an converges to a sum A.

    So, for [tex]\frac{(-1)^n}{\sqrt{n}}[/tex], we can write it as [tex](-1)^n \frac{1}{\sqrt{n}}[/tex].

    Since [tex]\left\langle \frac{1}{\sqrt{n}} \right\rangle[/tex] is a decreasing sequance (& also divergent), the series converge by virtue of the alternating series test.

    Is this correct? Does it make any sense now?
  9. Feb 1, 2009 #8
    Makes sense to me, as an extreme example of why your original logic doesn't work, think about what the following will produce:

    [tex] \sum_{n=1}^{\infty} \frac{(-1)^n}{n} [/tex] vs. [tex]\sum_{n=1}^{\infty} \frac{(1)^n}{n} [/tex]

    The fact that you have alternating terms is a pretty big deal.
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