1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    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


    User Avatar
    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook