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Real Analysis: Weierstrass M-Test

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that the series [tex]\sum[/tex](-1)n(x2+n)/n2 is uniformly convergent on every bounded interval in R, but is not absolutely convergent for any x.

    2. Relevant equations
    Weierstrass M-Test

    3. The attempt at a solution
    Take g_n(x) = (-1)n(x2+n)/n2. Then |g_n(x)| = (x2+n)/n2. To be uniformly convergent on every bounded interval in R, that means it would have to be uniformly convergent on any I = [a, b] s.t. a, b [tex]\in[/tex] R. So now I need to find an Mn such that each Mn >= 0, |g_n(x)| <= Mn, and [tex]\sum[/tex]Mn converges.

    But for any x, [tex]\sum[/tex]|g_n(x)| >= [tex]\sum[/tex]1/n, which does not converge. What is going on here? Thanks for any help.
  2. jcsd
  3. Oct 1, 2008 #2
    You have shown that it is not absolutely convergent by showing that |g_n(x)| does not converge. To show uniform convergence, you need to show that for every [tex]\varepsilon > 0[/tex] there is an N such that for all [tex]n\geq N[/tex], [tex]|g_n(x) - g(x)| < \varepsilon[/tex], where g(x) is the limit.

    Edit: I don't believe the Weierstrass M test will work for this.
    Last edited: Oct 1, 2008
  4. Oct 1, 2008 #3
    Thanks for your response, cellotim.

    How about if I use the fact that (g_n) converges uniformly to g on I = (a, b) iff lim(sup|g(x) - g_n(x)|) = 0, where x is in I.

    So sup|g(x) - g_n(x)| = sup(x2 + n)/n2 = (b2 + n)/n2. So lim(sup{(b2 + n)/n2}) = 0, and the series is uniformly convergent for every bounded interval in R.
  5. Oct 1, 2008 #4


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    There's a very easy way to estimate the error you make by truncating a series if it's an alternating series. Why not use that?
  6. Oct 1, 2008 #5
    I'm not sure why what I did was incorrect. I'm not really sure what you're getting at here. Sorry. :redface: What do you mean by "estimate the error you made"?
  7. Oct 1, 2008 #6


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    The series converges uniformly if the sequence of partial sums converge uniformly. So if you define g_n(x) to be the nth partial sum, sure, you want to show you can make |g_n(x)-g(x)|<epsilon. But |g_n(x)-g(x)| isn't a single term, it's an infinite series of all the terms in g(x) after the nth term. It the error you make by ignoring all those terms. In the case of an alternating series there is an easy way to estimate that error. Try Googling alternating series and see if you can find it.
  8. Oct 1, 2008 #7
    OK, let's see if this works.

    Define an = (x2 + n)/n2. lim(an) = 0, so the series [tex]\sum[/tex](-1)nan converges. By lim(an) = 0, there exists N such that n > N implies an < [tex]\epsilon[/tex].

    Then |[tex]\sum_{n=0}^\infty[/tex](-1)nan - [tex]\sum_{n=0}^N[/tex](-1)nan| < aN+1 < [tex]\epsilon[/tex]. So the series converges uniformly.

    Is this right? Thanks for your help.
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