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Here is a question that confused me much. Actually I can see the bijection, but don't know how to prove it.enter image description here

Here is a theorem that might useful, http://amininima.wordpress.com/2013/05/08/conditional-convergence/ but it uses bijection as a known factor, to prove the convergence of the second series. Could anyone help me? Thanks everyone in advance.

asaak
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  • You will find your answers here: https://en.wikipedia.org/wiki/Riemann_series_theorem – vadim123 Sep 30 '13 at 22:14
  • This seems obviously false for a complex sequence. If all the $a_n$'s are "pure imaginary", then the sum of them in any order, if it converges, will be "pure imaginary", which excludes most complex numbers, including all nonzero reals. – Stefan Smith Oct 01 '13 at 02:39

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To provide a more intuitive answer than the comment pointing to Riemann series theorem, this is a good example of shuffling the big stuff off to infinity. So, you know that $\Sigma|a_n| = \infty$, which also means that $\Sigma \min(a_n,0) = -\infty$ and $\Sigma \max(a_n,0) = \infty$

Given, however that the actual sum converges, the positives and negatives have to get smaller and smaller.

So, say you want the answer to be $r$, you add up the appropriately-signed components until you get $r$, then, using up the largest positive negative and positive pieces, go a little over and a little under using up the rest of the summands until they're all gone (or you can transition to the limit).

Chris Bonnell
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