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Convergence and uniformly convergence question

  1. Dec 24, 2008 #1
    1. The problem statement, all variables and given/known data
    For what range of positive values of x is [tex]\sum_{n=0}^\infty \frac{1}{1+x^n}[/tex]
    (a) convergent
    (b) uniformly convergent

    2. Relevant equations


    3. The attempt at a solution

    I didn't figure out how to separate convergence and uniformly convergence for this series.
    My idea was to consider two different intervals: x in [0,1] and in (1,[tex]\infty[/tex]).
    For the first interval,
    [tex]\frac{1}{1+x}+\frac{1}{1+x^2}+\cdots + \frac{1}{1+x^n}\ge \frac{n}{1+1}[/tex];
    and therefore, is divergent.
    For the other one, I considered a similar idea,
    [tex]\frac{1}{1+x}+\frac{1}{1+x^2}+\cdots + \frac{1}{1+x^n}\le \frac{n}{1+x}[/tex];
    but for this series be convergent, is necessary that x > n.
    So, please, can anyone give a little help here?
    How do I decide about the convergence? And uniform convergence?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 24, 2008 #2

    lurflurf

    User Avatar
    Homework Helper

    your second comparison is not helpful, try instead
    1/(1+x)+1/(1+x^2)+...+1/(1+x^n)<1/x+1/x^2+...+1/x^n=(x^(n-1)-1)/(x^n-x^(n-1))

    It makes not sense to compare x and n the way you do, because n takes an infinite number of values.

    Convergence is uniform if N can be chosen without regard to x.
     
  4. Dec 26, 2008 #3
    Thanks, lurflurf. Your idea was of a great help. But the uniform convergence is still giving me headache. I was thinking that if I could prove that the sum [tex]\sum \frac{1}{1+x^n}[/tex] is a monotonically decreasing function, for any value of n, then the bigger the x, smaller the sum, so It would be necessary less terms for establishing the convergence, and this way, this series wouldn't be uniformly convergent. What do you think?
     
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