Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculus II - Power Series

  1. Aug 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the radius of convergence and the interval of convergence of the series sigma[n=1,inf] ((3x-2)^n/(n^2*3^n))

    2. Relevant equations

    3. The attempt at a solution

    sigma[n=1,inf] ((3x-2)^n/(n^2*3^n))
    I applied the Root Test
    p=lim n->inf |(3x-2)^n/(n^2*3^n)|^(1/n) = lim n->inf |(3x-2)/(3n^(2/n))| = 0
    So the series Converges but I'm lost as to how to come up with the radius of convergence or the interval of convergence
    Last edited: Aug 20, 2011
  2. jcsd
  3. Aug 20, 2011 #2
    My bad I did this completley wrong let me work it out some more
  4. Aug 20, 2011 #3
    p=lim n->inf |(3x-2)^n/(n^2*3^n)|^(1/n) = lim n->inf |(3x-2)/(3n^(2/n))| = |x-2/3|

    sense for the root test 0<= p < 1
    I have
    0<= |x-2/3| <1
    whose solution is
    -1/3 < x < 5/3
    which is my interval of convergence
    So the radius of convergence is
    (5/3-1/3)/2 = 2/3

    does this look better?
  5. Aug 20, 2011 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Looks okay except you made a small error calculating the radius of convergence. You should find it equal 1.
  6. Aug 21, 2011 #5
    Ah thanks
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook