Series unchanged by rearrangement implies absolute convergence? If a series converges absolutely, then it is known that the value of the series is independent of rearrangements. More precisely, if $\sum |a_n| < \infty$ and $\sigma:\Bbb N\to \Bbb N$ is a bijection then $\sum a_{\sigma(n)}$ converges and its value is independent of $\sigma$.
Now what about the converse? That is,  if $\{a_n\}_{n=1}^\infty$ is an arbitrary real or complex sequence such that for every bijection $\sigma$ of $\Bbb N$, we have $\sum a_{\sigma(n)}$ converges to the same value,  is it true that $\sum a_n$ coverges absolutely? I am interested in this in the context of signed measures on a measurable space à la Stein's and Shakarchi's definition of signed measures in their third volume on real analysis.
 A: Based on Daniel Fischer's comments, we can prove the converse is true by using the contrapositive. Breaking the statement down in the case of real series, we are asserting that if there exists $\alpha\in\Bbb R$ such that for every permutation $\sigma$ we have $\sum a_{\sigma(n)}$ converges and $\sum a_{\sigma(n)} = \alpha$, then $\sum a_n$ converges absolutely. The kind of convergence in the hypothesis is known as "unconditional convergence."
Thus, the contrapositive of the statement we're looking for is "if $\sum a_n$ is not absolutely convergent, then it is not unconditionally convergent." 
If the series $\sum a_n$ is not absolutely convergent, then either $\sum a_n$ does not converge, or it converges, but not absolutely. In the first case, $\sum a_n$ is readily seen to be not unconditionally convergent, since unconditional convergence implies convergence in the usual sense. In the second case, $\sum a_n$ is convergent, but not absolutely, so the Riemann rearrangement theorem supplies two distinct permutations $\sigma,\tau$ such that $\sum a_{\sigma(n)},\sum a_{\tau(n)}$ both exist in the extended sense, possibly equal to $\pm\infty$, and they are unequal, so the series is not unconditionally convergent.
