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I have studied that for a sequence of real numbers absolute summability implies summability. What can we say about the sequences $\{x_k\}$ in a normed space . If it is not true in general could anybody show me with the counter examples in normed space?

Thanks for the help.

Srijan
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    @Martin Please don't close the question. I am beginner in functional analysis so whatever has written there I am not able to understand. If possible I need simple example that can make me understand. – Srijan May 25 '13 at 12:20
  • The question itself contains a simple example in $\ell_p$ with $p \gt 1$ which you should be able to understand. – Martin May 25 '13 at 12:22
  • @martin Series $\sum_{n=1}^\infty \frac{1}{n}e_n$ is unconditionally convergent but not absolutely convergent in $l_p$. If you could explain me how? – Srijan May 25 '13 at 12:28
  • Well, what is $\sum_{n=1}^\infty \lVert \frac1n e_n\rVert_p$? This will show that it is not absolutely convergent. To see that it is unconditionally convergent, show that for every strictly increasing sequence $(n_i)$ of natural numbers the sequence $\sum_{i=1}^\infty \frac{1}{n_i} e_{n_i}$ is Cauchy. – Martin May 25 '13 at 12:35
  • @Martin Thanks for the help. – Srijan May 25 '13 at 12:45
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    $\sum x_n$ is unconditionally convergent if and only if every subseries $\sum_{k=1}^\infty x_{n_k}$ converges. In $\ell_p$, $1<p<\infty$, for $x_n={1\over n} e_n$ this is the case. Note that $\sum x_n$ is just the vector $(1,1/2,\ldots)$. A subseries will just "zero out coordinates". E.g., $\sum_k x_{2k}=(0,1/2,0,1/4,\ldots)$. – David Mitra May 25 '13 at 12:47

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