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Cauchy sequences and absolutely convergent series

  1. Oct 18, 2014 #1

    Fredrik

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    1. The problem statement, all variables and given/known data

    I want to prove that if X is a normed space, the following statements are equivalent.

    (a) Every Cauchy sequence in X is convergent.
    (b) Every absolutely convergent series in X is convergent.

    I'm having difficulties with the implication (b) ⇒ (a).

    2. Relevant equations

    Only some standard definitions.

    3. The attempt at a solution

    I have no problems with the implication (a) ⇒ (b). I'm including the proof only because it can perhaps provide some inspiration for the proof of the converse.

    (a) ⇒(b): Let ##\sum_{k=1}^\infty x_k## be an arbitrary absolutely convergent series in X. For each ##n\in\mathbb Z^+##, define ##s_n=\sum_{k=1}^n x_k## and ##t_n=\sum_{k=1}^n \|x_k\|##. The desired result follows from
    $$\|s_n-s_m\|=\left\|\sum_{k=m+1}^n x_k\right\| \leq\sum_{k=m+1}^n\|x_k\| =|t_n-t_m|.$$ The argument is simple: ##(t_n)## is convergent and therefore Cauchy. This and the inequality above together imply that ##(s_n)## is Cauchy. Now (a) implies that ##(s_n)## is convergent.

    Here's the part I'm struggling with:

    (b) ⇒ (a): Let ##(s_n)## be an arbitrary Cauchy sequence in X. I'm using the notation ##s_n## rather than ##x_n## because I'm going to define a sequence ##(x_n)## such that ##(s_n)## is its sequence of partial sums. We define ##x_1=s_1## and for each ##n\geq 2##, ##x_n=s_n-s_{n-1}##. We have
    $$\sum_{k=1}^n x_k =(s_n-s_{n-1})+(s_{n-1}-s_{n-2})+\cdots+(s_2-s_1)+s_1 =s_n.$$ If we can show that ##\sum_{k=1}^\infty x_k## is absolutely convergent, we're done. Unfortunately that doesn't look possible. I think we have to continue something like this: For each ##k\in\mathbb Z^+##, let ##n_k## be a positive integer such that
    $$i,j\geq n_k\ \Rightarrow\ \|s_i-s_j\|<2^{-k}.$$ Let ##k\in\mathbb Z^+## be arbitrary. Define ##m_k=n_k+1##. We have
    $$\|x_{m_k}\|=\|s_{m_k}-s_{m_k-1}\|=\|s_{n_k+1}-s_{n_k}\|<2^{-k}.$$ Does this help at all? It tells us that ##\sum_{k=1}^\infty x_{m_k}## is absolutely convergent, and therefore convergent. But how does knowing that ##\sum_{k=1}^\infty x_{m_k}## is convergent help us? It means that the sequence ##\big(\sum_{k=1}^p x_{m_k}\big)_{p=1}^\infty## is convergent, but this isn't a subsequence of ##(s_n)##. ##(s_{m_p})_{p=1}^\infty## is a subsequence of ##(s_n)##, but I don't see why it would be convergent.
     
  2. jcsd
  3. Oct 18, 2014 #2

    PeroK

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    There's a proof here:

    https://www.math.ucdavis.edu/~thomases/201A_F10_hw3_sol.pdf [Broken]
     
    Last edited by a moderator: May 7, 2017
  4. Oct 18, 2014 #3

    Fredrik

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    Thanks. I understand it now. I need to write ##\|s_{n_{k+1}}-s_{n_k}\|<2^{-k}## instead of ##\|s_{n_k+1}-s_{n_k}\|<2^{-k}##. I tried that earlier, but somehow I convinced myself that if I do this, the sum fails to telescope.
     
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