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Uniform convergence of series

  1. Apr 8, 2009 #1
    Does [itex]\sum_{n=0}^{\infty} x^n(1-x)[/itex] converge uniformly on [itex](0,1)[/itex]?

    [itex]S_n(x)=\sum_{k=0}^{\infty} x^k(1-x)=\frac{1-x^{n+1}}{1-x}(1-x)=1-x^{n+1}} \rightarrow 1[/itex] as [itex]n \rightarrow \infty[/itex]
    so we get pointwise convergence to 1

    now we test for uniform convergence

    [itex]d_{\infty}(S_n,1) = sup |S_n(x)-1|= sup x^{n+1}[/itex] now remembering that x is in (0,1), i said that this should tend to 0 as n goes to infinity giving uniform convergence but the answers say that it doesn't go to 0 and so convergence is not uniform, merely pointwise - why is this?
     
  2. jcsd
  3. Apr 8, 2009 #2
    Re: Convergence

    Hint: this problem is screaming Weierstrass M-test.
     
  4. Apr 9, 2009 #3
    Re: Convergence

    we can find a sequence of positive integers [itex]M_n[/itex] such that [itex]|f_n(x) \leq M_n \forall x \in (0,1)[/itex] so let [itex]M_n=x^n[/itex] then since [itex]\sum M_n[/itex] converges, [itex]\sum f_n(x)[/itex] converges by Weierstrass M Test?

    why is it screaming M test?
     
  5. Apr 9, 2009 #4
    Re: Convergence

    It is screaming M test because that is what you have to use to answer the question.

    You can't use xn for Mn because Mn is independent of x. Try again.
     
  6. Apr 9, 2009 #5
    Re: Convergence

    just [itex]M_n=1 \forall n[/itex] then? this obviously converges and so we would get uniform convergence...?
     
  7. Apr 9, 2009 #6
    Re: Convergence

    Yes, Mn = 1 works and so by the M-test, you have uniform convergence on (0,1).
     
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