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Homework Help: Convergence on [-M,M] for any M implies convergence on R?

  1. Oct 30, 2011 #1
    Good day dear fellows. I am given the following series

    [itex] h(x) = \sum_{n=1}^{\infty} \frac{1}{x^2+n^2}. [/itex]

    It is requested that I show that [itex]h(x)[/itex] is continuous on R. I did the following: use the Weirerstrass M-test to show uniform convergence, and then, using the continuity of the functions that are being summed, conclude that [itex] h(x) [/itex] is continuous. The solutions manual agrees with me. So far so good.

    Next I am asked whether [itex] h(x) [/itex] is differentiable, and, if [itex] h(x) [/itex] is differentiable, to investigate if the derivative [itex] h'(x) [/itex] is continuous. I note that that

    [itex] \frac{d}{dx} \frac{1}{x^2+n^2} = \frac{-2x}{(x^2+n^2)^2}.[/itex]

    At this point I don't see how I am going to get uniform convergence out of a summation over this thing, so I take out the solutions manual again. The manual uses the Weirerstrass M-test to show uniform convergence on [itex] [-M,M] [/itex] for arbitrary [itex]M[/itex], and concludes that there must be uniform convergence on R (because M is aribitrary). To be explicit: on any interval [itex] [-M,M] [/itex],

    [itex] |\frac{2x}{x^2+n^2} | \leq \frac{2M}{n^2} [/itex].

    (the square over the denominator can be removed since the denominator is greater than one).

    And, since

    [itex] \sum_{n=1}^{infty} \frac{2M}{n^2} [/itex]

    converges on [-M,M], uniform convergence of the derivatives follows via the weirerstrass M-test. Since [itex]M[/itex] is arbitrary, we then have uniform convergence (and thus continuity) on R.

    I don't see how this is so, since R is not a closed interval..
  2. jcsd
  3. Oct 30, 2011 #2


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    Science Advisor

    On the contrary, R is a closed interval. It is just not a bounded interval.

    In any case, "convergence", in a set X, just means the series converges at every point of X. Let p be any point in R. Then there exist M such that -M< p< M. If the series converges on [-M, M] then it converges at p. Since p could be any point in R, the series converges for all R.

    "Uniform convergence" would be a different matter.
  4. Oct 30, 2011 #3
    I think I am confused by the notation for intervals. For any finite numbers [itex] a [/itex] and [itex]b [/itex], [itex] (a,b) [/itex] is an open interval, and every closed interval can be written [itex] [c,d], [/itex] where [itex]c[/itex] and [itex]d[/itex] are its endpoints. But this isn't true for unbounded intervals? In other words, [itex] (-\infty, \infty) [/itex] is a closed interval, since it contains all it's limit points (but not its endpoints)? Endpoints of unbounded intervals are not limit points?
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