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!

QUESTION: Interval of Convergens for a series

  1. May 10, 2006 #1

    I have this series here

    [tex]\sum_{n=1} ^{\infty} \frac{1}{x^2+n^2}[/tex]

    I need to show that the Radius of convergens [tex]R = \infty [/tex] and the interval of convergens therefore is [tex](- \infty, \infty) [/tex]

    My question is to do this don't I use the ratio-test?

    Sincerely Yours

    Last edited: May 10, 2006
  2. jcsd
  3. May 10, 2006 #2
    a_n is the nth term, so:

    [tex]\left| \frac{a_{n+1}}{a_n} \right | = \left| \frac{1}{x^2+(n+1)^2} \cdot \frac{x^2+n^2}{1} \right |[/tex]

    However, this goes to 1 as n goes to inifinity so this really doesn't help you. What you probably want to do is use comparison test with 1/n^2 to show that it converges for any x.
    Last edited: May 10, 2006
  4. May 10, 2006 #3
    Hi and thanks for Your answer,

    [tex]\sum_{n=1} ^{\infty} \frac{1}{x^2+n^2}[/tex]

    Then by the comparison test:

    [tex]\frac{1}{x^2 + n^2} < \frac{1}{n^2}[/tex] ??



  5. May 10, 2006 #4
    Thats right, and since [tex]\sum_{n=1} ^{\infty} \frac{1}{n^2}[/tex] converges (by p-series), [tex]\sum_{n=1} ^{\infty} \frac{1}{x^2+n^2}[/tex] converges since every term is smaller.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: QUESTION: Interval of Convergens for a series