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Homework Help: Power Series

  1. Nov 2, 2006 #1
    Could you check over my work on these problems:


    1. For what values of [tex] x [/tex] is the series [tex] \sum_{n=0}^{\infty} n!x^{n} [/tex] convergent?

    I used the ratio test: [tex] \lim_{n\rightarrow \infty} |\frac{a_{n+1}}{a_{n}}| = \lim_{n\rightarrow \infty}(n+1)|x| = \infty [/tex]. So it diverges by the ratio test, but by the definition of the power series it converges at [tex] x = 0 [/tex]


    2. For what values of [tex] x [/tex] does the series [tex] \sum_{n=1}^{\infty} \frac{(x-3)^{n}}{n} [/tex] converge? I again used the ratio test and came up with the following:

    [tex] \lim_{n\rightarrow \infty} \frac{1}{1+\frac{1}{n}}|x-3| = |x-3| [/tex]. Therefore the series diverges when [tex] |x-3|< 1 [/tex] or when [tex] 2 < x < 4 [/tex]. Testing the endpoints, it converges when [tex] 2\leq x< 4 [/tex]


    3. Find the radius of convergence and the interval of convergence of the series [tex] \sum_{n=0}^{\infty} \frac{n(x+2)^{n}}{3^{n+1}} [/tex]. Using the ratio test I came up with [tex] \frac{|x+2|}{3} [/tex]

    Thus [tex] |x+2|<3 \rightarrow R = 3 [/tex]. Testing at the endpoints, the interval of convergence is [tex] (-5, 1) [/tex]

    Thanks
     
    Last edited: Nov 2, 2006
  2. jcsd
  3. Nov 2, 2006 #2
    delete another thread of the same title please
     
  4. Nov 2, 2006 #3

    StatusX

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    Homework Helper

    Looks good.
     
  5. Nov 2, 2006 #4
    Thanks. One more question:

    A function [tex] f [/tex] is defined by [tex] f(x) = 1+2x+x^{2}+2x^{3}+x^{4}+... [/tex]. Find the interval of convergence, and an explicit formula for [tex] f(x) [/tex].

    So [tex] c_{2n} = 1, c_{2n+1} = 1 [/tex]. So [tex] f(x) = \sum_{n=0}^{\infty} x^{2n} + 2x^{2n+1} [/tex]. Then just apply the ratio test to this to find the interval of convergence?
     
    Last edited: Nov 2, 2006
  6. Nov 2, 2006 #5

    Hurkyl

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    A minor detail: you should indicate where the ratio test says it diverges.


    This is a job for the n-th root test! (In fact, once you're used to it, it's almost always at least as good as the ratio test)

    I think what you've done is okay... regrouping terms is always a tricky business, but I can you can apply these theorems:

    (1) A power series converges absolutely in the interior of its interval of convergence.

    (2) You can rearrange / regroup the terms of an absolutely convergent sequence in any way you want.
     
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