# The sum of an unconditionally summable sequence does not depend on the order of the terms

If I have a series of elements of a Banach space that converges unconditionally (it converges regardless of the order of the terms), why is it the case that the sum does not depend on the order of the terms?

I'm reading a text on bases in Banach spaces and it threw out "The sum does not depend on the permutation of the index set" as a remark, but I'm finding it more difficult than I expected to prove this (I'm a beginner at this sort of thing).

Thanks

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Do you understand this result for real numbers? – Phira Dec 5 '12 at 2:50
I'm not certain what you're asking. You seem to have posed a question of the form "If (thing) holds, then why does (thing) hold?" – Cameron Buie Dec 5 '12 at 2:55
I understand the result for absolutely convergent series in the real numbers, where absolutely=unconditionally, but the proof that I know of the result for the reals uses comparison test and the nice ordering properties of the reals. – user51823 Dec 5 '12 at 3:00
Cameron: The definition of unconditional convergence as given in the book was "It converges to [some sum] regardless of the order of the terms". I want to know why the sum is always the same. ie. Why can't we rearrange it to get a series that still converges (so the definition of unconditional convergence holds), but to a different sum than the original series converges to. – user51823 Dec 5 '12 at 3:03
Found a proof in Kadets' book: Series in Banach Spaces: Conditional and Unconditional Convergence. For anyone else who has the same question, here's a hint: Suppose they converge to two different sums S1 and S2 and consider a (bounded) functional f such that f(S1) is not equal to f(S2)... – user51823 Dec 5 '12 at 3:13