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Sketchy Induction Proof

  1. Feb 28, 2007 #1
    something funny's going on here, and I can't see what :yuck:

    For a sequence [tex] {x_n} [/tex] , where each term is non-negative

    the series [tex] x_1 + x_2 + ... +x_n + ... [/tex] converges


    it will suffice to show that the sequence of partial sums [tex] {s_n} [/tex] is bounded

    where each [tex]s_i = x_1 + ... + x_i [/tex]

    when i=1,
    [tex] s_1 = x_1 [/tex]

    so the result holds true for i=1

    let the result be true for all positive numbers up to some k such that

    [tex] s_k <= some b [/tex]

    now consider [tex] s_{k+1} [/tex].....

    [tex] s_{k+1} = s_k + x_{k+1} <= b + x_{k+1} [/tex]

    so the result holds true for all k= 1, 2, 3 ....
    Last edited: Feb 28, 2007
  2. jcsd
  3. Feb 28, 2007 #2
    Of course finite sums are always bounded!

    You're implicitly implying that there are infinitely many partial sums. Induction doesn't work for infinity, it only works for a given natural number.

    ie, you haven't shown that the supremum of all partial sums exists.
  4. Feb 28, 2007 #3
    no it won't

    You need to show: there exist a S such that for any given Epsilon greater than 0 there exist a N (element of the natural numbers) where |S_n - S| < Epsilon provided that n>N.
    Last edited: Feb 28, 2007
  5. Feb 28, 2007 #4
    Well, since the terms in the partial sums are non-negative all he has to show is that ALL the partial sums are bounded by some number. And then he has a bounded monotone sequence of partial sums which implies they converge.

    The induction showed that for every partial sum s_n there existed a number b such that s_n<b. This DOES NOT mean there exists an m such that s_n<=M for all n.
  6. Feb 28, 2007 #5
    ah, thanks for clearly that up. That's kind of subtle, and I doubt I would've figured it out by myself..

    i guess i forgot that one of the reasons why we speak of partial sums in the first place is that they are all bound..
  7. Mar 1, 2007 #6


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    Induction doesn't work that way! Induction on n shows that statement Sn is true for any finite n.
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