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Convergence of series of functions

  1. May 7, 2005 #1
    Hi all,
    I have few questions about this excercise:

    Analyse pointwise, uniform and local uniform convergence of this series of functions:

    [tex]
    \sum_{k=2}^{\infty}\log \left(1 + \frac{x^2}{k \log^2 k} \right)
    [/tex]

    I'm trying to do it using Weierstrass' criterion. To recall it, it says

    [tex]
    \mbox{Let } f_n \mbox{ are defined on } 0 \neq M \subset \mathbb{R}\mbox{, let }
    S_n := \sup_{x \in M} \left| f_{n}(x)\right|, n \in \mathbb{N}. \mbox{ If }
    \sum_{n=1}^{\infty} S_n < \infty\mbox{, then } \sum_{n=1}^{\infty} f_{n}(x) \rightrightarrows \mbox{ on } M.
    [/tex]

    How to find

    [tex]
    \sup_{x \in M} \left| f_{n}(x)\right|
    [/tex]
    ?

    The derivative is
    [tex]
    \left(\log \left(1 + \frac{x^2}{k \log^2 k} \right)\right)^{'} = \frac{2x}{k\log^2 k + x^2}
    [/tex]

    It means that the function is growing for [itex]x > 0[/itex]. x going to infinity would bring us problems, so I will take [itex]x \in [-K, K][/itex], where [itex]-\infty < -K < K < \infty[/itex].

    Then
    [tex]
    \sup_{x \in M} \left| f_{n}(x)\right| = \log \left( 1 + \frac{K^2}{n \log^{2} n}\right)
    [/tex]

    But I don't know how to prove that
    [tex]
    \sum_{n=2}^{\infty} \log \left( 1 + \frac{K^2}{n \log^{2} n}\right) \mbox{ converges}
    [/tex]

    Could someone point me to the right direction please?

    Thank you.
     
  2. jcsd
  3. May 7, 2005 #2

    quasar987

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    You don't necessarily have to find the sequence of suprema. If you find a sequence of upper bounds that converges, then according to the "regular" comparison criterion, the serie of suprema does too!

    In other words, if you find a sequence [itex]a_n[/itex] such that [itex]|f_n(x)| \leq a_n[/itex] [itex]\forall x \in M[/itex] and at least for n>N, and such that [itex]\sum a_n < \infty[/itex], then since [itex]S_n \leq a_n[/itex] (definition of supremum), we have [itex]\sum S_n < \infty[/itex] (comparison criterion for numerical series)
     
  4. May 7, 2005 #3
    Thank you quasar987, I have already also thought of comparing it with some convergent series, but I haven't thought any proper up. Now I'm trying to solve the problem using the theorem about change of sum and derivation and it may help...
     
  5. May 7, 2005 #4

    quasar987

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    I think I found something but you'd better double check. I proved that the serie does not converge pointwise.

    1) [itex]Dom f_n = \mathbb{R}[/itex].

    2) Consider an element [itex]x_0 \in \mathbb{R}[/itex].

    3) Then, according to a theorem for numerical series,

    [tex]\sum f_n(x_0) \ \ \mbox{converges} \Leftrightarrow \sum 2^n f_{2^n}(x_0) \ \ \mbox{converges}[/tex]

    Now calulate [itex]2^n f_{2^n}(x_0)[/itex] and evaluate the limit as n goes to infinity (you'll need to use l'Hospital's rule once). I find that the result is infinity. But according to a theorem for numerical series, if [itex]\sum a_n \neq 0 [/itex], it diverges. So our serie of function does not converge pointwise for any element of R.. ==> it does not converge uniformally on any interval.

    But like I said, double check this, it seems kinda dubious.

    (what does local convergence mean?)
     
    Last edited: May 7, 2005
  6. May 7, 2005 #5
    Well, first I'll show you how I tried:

    We have theorem saying this:

    [tex]
    \mbox{Let } f_n, n \in \mathbb{N} \mbox{ are defined and have finite derivatives } f_{n}^{'} \mbox{ on } (a, b) \subset \mathbb{R}. \mbox{ Let \\}
    [/tex]

    [tex]
    \mbox{\\ (i) } \sum_{n=1}^{\infty} f_{n}^{'} \stackrel{loc}{\rightrightarrows} \mbox{ on } (a,b)
    [/tex]

    [tex]
    \mbox{ \\ (ii) } \exists x_0 \in (a,b) \mbox{ such, that } \sum_{n=1}^{\infty} f_{n}(x_0) < \infty \mbox{ (converges) \\}
    [/tex]

    [tex]
    \mbox{ \\ Then } \sum f_{n} \stackrel{loc}{\rightrightarrows} \mbox{ on } (a,b).
    [/tex]

    So I have

    [tex]
    f_{n}^{'} = \frac{2x}{n\log^{2}n + x^2}
    [/tex]

    Now this is the function for which we want to analyse uniform convergence. Where's the supremum? Derivative of derivative is

    [tex]
    \frac{2n\log^{2}n - 2x^2}{\left(n\log^{2}n + x^2\right)^2} = 0 \Leftrightarrow x = \pm \sqrt{n\log^{2}n}
    [/tex]

    So the supremum is in [itex]x = \sqrt{n\log^{2}n}[/itex], however, because [itex]x \in [-K,K][/itex] (it won't move along with growing n), supremum will move to the rightmost point of the interval. So we have

    [tex]
    \sup_{x \in [-K,K]} \left| \frac{2x}{n\log^{2}n + x^2}\right| = \frac{2K}{n\log^{2}n + K^2} =: S_n
    [/tex]

    Then
    [tex]
    \sum S_n < \infty \Rightarrow \sum f_{n}(x) \mbox{ converges on} [-K, K]
    [/tex]

    According to results it's ok, I just want to ask you if you think it's mathematically correct (ie. ok to write it this way in test)
     
    Last edited: May 8, 2005
  7. May 7, 2005 #6

    quasar987

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    What's the definition of local convergence?

    I'm not one to answer that. I'm just learning this stuff, like you.
     
  8. May 7, 2005 #7
    [tex]
    \mbox{We say that } \left\{ f_{n}\right\}_{n=1}^{\infty} \mbox{ converges locally uniformly to f on M and write}
    [/tex]

    [tex]
    f_{n} \overset{loc}{\rightrightarrows} f \mbox{ on M, if}
    [/tex]

    [tex]
    \mbox{for each } x \in M \mbox{ there exists } \delta > 0 \mbox{ such, that } \left\{f_{n}\right\}_{n=1}^{\infty} \mbox{ converges uniformly to } f \mbox{ on } M\ \cap \ U(x, \delta).
    [/tex]
     
  9. May 7, 2005 #8

    quasar987

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    [tex]
    f_{n}^{'} = \frac{2x}{n\log^{2}n + x^2} = 0 \Leftrightarrow x = \pm \sqrt{n\log^{2}n}
    [/tex]


    How do you get this? I would agree that (for n fixed), f ' = 0 <==> x = 0.
     
  10. May 8, 2005 #9
    A typo, I corrected it in my post.
     
    Last edited: May 8, 2005
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