17
$\begingroup$

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.

$\endgroup$
5

1 Answer 1

Reset to default
26
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Why can we find a subsequence $n_k$ such that $|x_{n_{k+1}}-x_{n_k}|<2^{-k}$? $\endgroup$
    – Sujaan Kunalan
    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. $\endgroup$
    – Emanuele Paolini
    Oct 1, 2014 at 20:15
  • 2
    $\begingroup$ 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
    May 2, 2015 at 12:48
  • $\begingroup$ You are right! I have corrected it.... $\endgroup$
    – Emanuele Paolini
    May 2, 2015 at 14:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .