# About Banach Spaces And Absolute Convergence Of Series

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.

• A complete proof can be found here: planetmath.org/… Feb 9, 2013 at 9:06
• So nice of you! Feb 23, 2013 at 8:26
– Mark
Oct 16, 2017 at 1:25
• Proof of 1 can be found here: proofwiki.org/wiki/Absolutely_Convergent_Series_is_Convergent
– Mark
Oct 16, 2017 at 1:46
• Serios' planetmath working link: planetmath.org/… Feb 24, 2020 at 11:46

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.
• Why can we find a subsequence $n_k$ such that $|x_{n_{k+1}}-x_{n_k}|<2^{-k}$? Oct 1, 2014 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. Oct 1, 2014 at 20:15
• May I ask why $\sum y_k = lim x_{n_k}$ instead of $\sum y_k = lim x_{n_k} - x_{n_1}$? May 2, 2015 at 12:48