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How to prove the following two assertions:

  1. If in a normed space $X$, absolute convergence of any series always implies convergence of that series, then $X$ is a Banach space.

  2. In a Banach space, absolute convergence of any series always implies convergence of that series.

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1 Answer 1

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Take a Cauchy sequence $x_k$. Then you can find a subsequence $n_k$ such that $|x_{n_{k+1}}-x_{n_k}| < 2^{-k}$. Let $y_k = x_{n_{k+1}}-x_{n_k}$. Then $\sum y_k$ is absolutely convergent and, by hypothesys, it converges. But $\sum_{k=1}^N y_k = x_{n_N} - x_{n_1}$ and hence $x_k$ has a convergent subsequence. But since $x_k$ is Cauchy, the whole sequence is convergent.

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    $\begingroup$ Why can we find a subsequence $n_k$ such that $|x_{n_{k+1}}-x_{n_k}|<2^{-k}$? $\endgroup$ Commented Oct 1, 2014 at 15:13
  • $\begingroup$ Choose $n_k$ so that $|x_m - x_{n_k}|<2^{-k}$ for all $m>n_k$. This is possible by the definition of Cauchy sequence. $\endgroup$ Commented Oct 1, 2014 at 20:15
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    $\begingroup$ May I ask why $\sum y_k = lim x_{n_k}$ instead of $\sum y_k = lim x_{n_k} - x_{n_1}$? $\endgroup$
    – TH000
    Commented May 2, 2015 at 12:48
  • $\begingroup$ You are right! I have corrected it.... $\endgroup$ Commented May 2, 2015 at 14:27

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