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Uniform convergence

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider

    [tex]f(x)=\sum\frac{1}{n(1+nx^{2})}[/tex]

    from n=1 to n=infinity.

    On what intervals of the form (a,b) does the series converge uniformly? On what intervals of the form (a,b) does the series fail to converge uniformly?


    2. Relevant equations

    Weierstrass M-test: If there exists numbers [tex]M_{r}[/tex] for each [tex]f_{r}(x)[/tex] such that [tex]f_{r}(x) \leq M_{r}[/tex] and

    [tex]\sum M_{r}[/tex]

    converges, then [tex]\sum f_{r}[/tex] converges uniformly.

    3. The attempt at a solution

    Write

    [tex]f_{r}(x)=\frac{1}{r(1+rx^{2})}[/tex] .

    If i just consider the case where x>1 then

    [tex]\frac{1}{r(1+rx^{2})} \leq \frac{1}{r^{2}x^{2}} \leq \frac{1}{r^{2}}[/tex]

    so let

    [tex]M_{r}=\frac{1}{r^{2}}[/tex]

    and the sum of this from r=1 to r=infinity converges. So

    [tex]\sum f_{r}[/tex]

    converges uniformly for x>1? Or am I misunderstanding the M-test?

    But then say I went with x>0.5 instead of x>1. I could then choose

    [tex]M_{r}=\frac{2}{r^{2}}[/tex]

    to get uniform convergence for x>0.5? etc etc

    So what's the required domain? I'm really confused :-(
     
  2. jcsd
  3. Nov 18, 2009 #2
    That's exactly correct if you just use M = 4/r^2 instead. How about uniform convergence for x>a where a is an arbitrary positive real number? You can generalise your answer easily. Then there's only one real number you're leaving out, and it turns out that the series does not converge there.
     
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