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Rudin Theorem 3.23

  1. Jul 12, 2012 #1
    Theorem 3.23 is a very simple one: it says that if a series converges then the limit of the terms of the sequence is zero. However Rudin's way of justifying this fact doesn't seem valid to me. He uses the following logic:

    A series converges if and only if the sequence of partial sums is cauchy meaning that for all ε > 0 there is an integer N s.t. for all n,m > N and n <= m the sum of the terms of the sequence from a_n to a_m is less than ε.

    Rudin says that the case where n = m proves this theorem. However when n = m the only thing the cauchy criterion states is that the distance from a_n to a_n approaches zero. It does not actually say that the value of a_n approaches zero.

    To prove this we need the case where n = m - 1.

    Then the difference between the two partial sums is a_m and therefore a_m approaches zero.

    Am I wrong?
     
  2. jcsd
  3. Jul 12, 2012 #2

    Yes, I think you are. In theorem 3.22, Rudin proved that
    [tex]\sum_{n=1}^\infty a_n\,\,\text{converges iff}\,\,\forall\epsilon >0\,\,\exists N\in \Bbb N\,\,s.t.\,\,\left|\sum_{k=n}^m a_k\right|<\epsilon\,\,\,\text{whenever}\,\,m\geq n\geq N [/tex]
    If we accept this (and I can't see any reason *not* to accept it), then we can take [itex]\,n=m\,[/itex] , from where we get that if the series converges then
    [tex]\forall \epsilon >0\,\,\exists\,N\in\Bbb N\,\,s.t.\,\,\left|\sum_{k=n}^n a_k\right|=|a_n|<\epsilon\,\,\,\text{whenever}\,\,\,n\geq N[/tex] which is exactly the definition of "the sequence [itex]\,\{a_n\}\,[/itex] converges to zero".

    Please do note that in the last part above we do NOT have the "iff" of theorem 3.22, since we're limiting [itex]\,m=n\,[/itex], yet the necessary part, of course, remains valid.

    DonAntonio
     
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