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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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A complete proof can be found here:… – Seirios Feb 9 '13 at 9:06
So nice of you! – Saaqib Mahmuud Feb 23 '13 at 8:26
up vote 13 down vote accepted

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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Why can we find a subsequence $n_k$ such that $|x_{n_{k+1}}-x_{n_k}|<2^{-k}$? – Sujaan Kunalan Oct 1 '14 at 15:13
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. – Emanuele Paolini Oct 1 '14 at 20:15
May I ask why $\sum y_k = lim x_{n_k}$ instead of $\sum y_k = lim x_{n_k} - x_{n_1}$? – TH000 May 2 '15 at 12:48
You are right! I have corrected it.... – Emanuele Paolini May 2 '15 at 14:27

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