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Radius and Interval of convergence

  1. May 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Ʃ(((-1)^n)(x^n))/(n+1) from 0 to ∞


    2. Relevant equations



    3. The attempt at a solution
    I took applied the ratio test and got that lim|(x^(n+1))/(n+2) * (n+1)/(x^n)| =|x|
    so that means for it to converge |x|<1 Radius of convergence is 1
    My interval is (-1<x<1)
    Now I check the endpoints

    x=1
    Ʃ(((-1)^n))/(n+1) from 0 to ∞
    I want to use the alternating series test, but I have run into a little problem.
    I know that the Lim 1/(n+1) =0 but how do I prove that 1/(n+2) < 1/(n+1) for all n.
    I know its true, but my professor wants us to show that. I can show that -1<1 which is a true statement, but what does that say about the inequality in relation to n?

    Is there a better test to do this? Ratio test just equals 1 so that tells me nothing. I can't really use the integral test for an alternating series since it would end up imaginary. If there is a better test could someone still show how to use the alternating series if possible?
     
  2. jcsd
  3. May 11, 2013 #2

    vela

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    Usually you can assume certain properties of real numbers to be true. If your professor really expects you to prove such a trivial point, why don't you start with something "obviously" true, like n+2 > n+1 for all n?
     
  4. May 11, 2013 #3

    HallsofIvy

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    Or, to start from the very obvious 2> 1. Add n to both sides, then divide both sides by ...
     
  5. May 12, 2013 #4
    I set 1/(n+1) to a function f which I then siad was less than 0. when i took the derivative of f i obtained a negative function which is less than 0. I stated that since the derivative is less than zero then the function is decreasing and bn+1 < bn. Is that ok?
     
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