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Riemann Series Theorem

  1. Apr 22, 2008 #1

    daniel_i_l

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    Can anyone tell we how this:
    http://mathworld.wolfram.com/RiemannSeriesTheorem.html
    can be proved?
    The book that I read it in said that it was "beyond the scope of the book".
    It one of the coolest theorems I've read about. For example, it means that for any number (pi, phi, ...) there's some series which converges to it.
     
  2. jcsd
  3. Apr 22, 2008 #2

    Hurkyl

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    The easiest way to see the proof is to simply try to do an example. Take the series whose positive terms are

    1, 1/3, 1/5, 1/7, ...

    and whose negative terms are

    -1/2, -1/4, -1/6, ...

    and see if you can choose 10 terms that add up to something near 2. Then try to extend it to 20 terms that add up to something even closer to 2, and so forth.
     
  4. Apr 22, 2008 #3

    daniel_i_l

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    Well I see that since we have an infinite amount of positive and negative terms we should be able to "work out" some balance between them that converges to any number. But how can you prove it?
    Thanks.
     
  5. Apr 22, 2008 #4

    Hurkyl

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    I, personally, would write an algorithm that generates the sequence term by term, and then prove the result has the properties I want.
     
  6. Apr 22, 2008 #5

    HallsofIvy

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    The proof is not difficult. Let {an} be a sequence that is convergent but not absolutely convergent. That is, [itex]a_1+ a_2+ a_3+ \cdot\cdot\cdot [/itex] converges but [itex]|a_1|+ |a_2|+ |a_3|+ \cdot\cdot\cdot[/itex] does not.

    Define [itex]b_n= a_n[/itex] if [itex]a_n\ge 0[/itex], 0 if not.
    Define [itex]c_n= -a_n[/itex] if [itex]a_n< 0[itex], 0 if not.


    For all n, [itex]a_n= b_n- c_n[/itex], [itex]|a_n|= b_n+ c_n[/itex]. In each of those, one term is 0, the other may not be.

    (In what follows "{xn} converges" means the series converges.)

    Suppose {bn} converges. Then [itex]c_n= b_n- a_n[/itex]. Since both {bn} and {an} converge, so does {cn}. But then {|an} must converge which is not true. Therefore, {bn} cannot converge. You can do the same thing to show that the series {cn} does not converge. Since the both consists of non-negative numbers, the partial sums must go to infinity.

    Let "a" be any real number. Then there exist n1 such that [itex]\sum1^{n_1} b_n[/itex]> a[/itex]. Let a1 be that sum minus a. Then there exist n2 so that [itex]\sum_1^{n_2} c_n> a_1[/itex]. The sum of the corresponding terms of {an}, with the correct sign, will be slightly less than a. Let a_2 be a- that number. There exist n3 so that [itex]\sum_{n_1}^{n_3} b_n> a_2[/itex]. Continuing in that way, we get a sequence of numbers from {an}, rearranged whose partial sums "alternate" on either side of a and converge to a.

    It's not too hard to see how to choose terms so the series diverges to +infinity or to -infinity.
     
  7. May 4, 2008 #6

    daniel_i_l

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    Thanks for posting that. But how do you know that the partial sums converge to a? In other words, how can you be sure that |a_1| > |a_2| > |a_3| ... ?
    Thanks.
     
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