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Two problems which have been giving me problems

  1. Feb 29, 2008 #1
    1. The problem statement, all variables and given/known data
    1) Test the following series for Uniform Convergence
    \sum\limits_{n = 1}^{\inf } {\frac{{( - 1)^n }}{{n^{x}\ln (x)}}}

    2) Let f(n,x) = [tex]
    \sum\limits_{n = 1}^{\inf } {( - 1)^n (1-x^{2})x^{n}}
    a) Test for absolutely convergence on [0,1]
    b) Test for uniformly convergence on [0,1]
    c) Is [tex]
    \sum\limits_{n = 1}^{\inf } {|f(n,x)|}
    [/tex] absolutely convergent on [0,1]?

    2. Relevant equations

    3. The attempt at a solution

    For the first, I'm utterly lost. Is there an easy way to deal with such series?

    For the second, could I just argue that for all 0<=x<1, there exists a, s.t. x < a <1
    and thus
    |\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}
    and for x = 1 and any a > 0
    |\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} = 0 < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}[/tex]

    This would prove all 3 right? But can I argue taht way? Can I fix my 'x' ahead of time, or does my argument have to work for all x simultaneously? Cuz if it does, then all I would have to do is choose x between a and 1 and the argument would break down.
  2. jcsd
  3. Feb 29, 2008 #2
    I think I figured out two (will post my work later).

    But for 1, could I argue that the sequences of partial sums cannot be uniformly cauchy since they are unbounded near x = 1.
  4. Feb 29, 2008 #3
    I confess I might have jumped the gun in saying that I figure out number two. Any help on either would still be greatly appreciated (also, if I could get a mod to give this thread a more decriptive title.... I don't think I can change the title).
  5. Feb 29, 2008 #4
    For the first one, how about splitting the sum into two: one for n even, one for n odd. Then you can use the p-series test on each.

    The argument you use when 0 <= x < 1 works. When x = 1, what's inside the sum is 0 so the whole sum is 0. I find it unnecessary to consider the absolute value of the sum.
  6. Mar 1, 2008 #5
    Call me stupid (I'd rather you don't though =P), but I'm not sure I understand your approach for the first.

    What I did was show that I can get x arbitrarily close to 1, and since log(x) is continuous, I can get arbitrarily close 1. Basically, I showed that for any 'n', I can make [tex]
    log(x) < \frac{1}{ne}
    so for x sufficiently close to 1

    |\frac{1}{{n^{x}\ln (x)}}| > \frac{1}{|{n||\ln (x)|}} > |\frac{ne}{n}| = e

    where first inequality holds because x is between 0 and 1.

    Therefore the terms of the series do not go to zero uniformly on its domain, so it can't converge.

    Is that a valid argument?
  7. Mar 1, 2008 #6
    What I wrote is for testing convergence, not uniform convergence. Sorry about that. I'm not familiar with uniform convergence.
  8. Mar 1, 2008 #7
    No problemo. =)

    (Btw, uniform convergence is essentially the same, but it has to work for arbitrary x in the interval).
  9. Mar 1, 2008 #8
    If x is a fixed constant greater than 0, by the alternating series test, the sum in 1) converges. Does this mean that the sum converges uniformly in the interval [itex](0, \infty)[/itex]?
  10. Mar 1, 2008 #9
    Oops, I forgot to say that x belongs to [0,1].

    Note that log(1) = 0, so it can't converge.

    To converge uniformly, it has to converge for every x in the interval.
  11. Mar 1, 2008 #10
    Please, if a mod could, edit the first one so x belongs to the closed interval [0,1].

    Thanks =)
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