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We have, from real analysis, that: Let $a_n$ be an array in $\mathbb{R}$, such that $\sum_{n=0}^{+\infty} |a_n|$ converges, then, for any bijection $\phi:\mathbb{N} \to \mathbb{N}$ it holds $\sum_{n=1}^{+\infty} a_{\phi(n)} = \sum_{n=1}^{+\infty} a_n$(we call this part of Riemann theorem, the other part states that if $\sum_{n=1}^{+\infty} |a_n| does not converge, you can obtain any real number by permuting).

Motivated by this, I do not see why it wouldn't hold a generalisation of this:

Let $(X, ||\cdot||)$ be a Banach space, and $a_n$ such an array that $\sum_{n=1}^{+\infty} ||a_n||$ is finite, then for any bijection $\phi:\mathbb{N} \to \mathbb{N}$ it holds $\sum_{n=1}^{+\infty} a_n = \sum_{n=1}^{+\infty} a_{\phi(n)}$.

Does this really hold? Thank you!

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    According to https://en.wikipedia.org/wiki/Absolute_convergence#Rearrangements_and_unconditional_convergence, absolute convergence implies unconditional convergence in a complete Banach space.2017-02-19
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    What you call "array" is usually known as "series" ...2017-02-19

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