Any rearrangement of an absolutely convergent sequence $(a_n)$ is another absolutely convergent sequence with the same limit. Let $(a_{\sigma(n)})$ be the rearranged sequence under the bijection of indices $\sigma$.

My proof attempt. Let $T_N = \sum_{n=1}^N |a_{\sigma(n)}|$ be the $N$th partial sum of the rearranged absolute series. Similarly let $S_N$ be the partial sum of the original absolute series. I know I want to try to prove the absolute sequences Cauchy. Still working on it.

  • $\begingroup$ So, not much of an attempt, then? $\endgroup$ – Andrés E. Caicedo Oct 5 '13 at 23:06
  • $\begingroup$ Hold on, I just thought of something $\endgroup$ – AlgebraicGeometryStudent Oct 5 '13 at 23:07
  • $\begingroup$ Here is the key hint: Given $n$, show that if $N$ is large enough, then all surviving terms in $T_N-S_N$ have index larger than $n$. $\endgroup$ – Andrés E. Caicedo Oct 5 '13 at 23:07
  • $\begingroup$ In think your first two "sequence" in the body of the question must be changed to "series". $\endgroup$ – DonAntonio Oct 5 '13 at 23:15
  • $\begingroup$ @DonAntonio The OP should rather say "absolutely summable". $\endgroup$ – Pedro Tamaroff Oct 5 '13 at 23:16

You needn't make the $N$s match. You know you can take $N$ large enough so that $$\sum_{k\geqslant N}|a_k|<\varepsilon$$

Now pick $M$ large enough so that all the terms in $T_M$ are some of the terms in $S_N$. Then how large is $$\left|\sum_{k\geqslant 0}a_k-T_M\right|?$$

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  • $\begingroup$ I don't know the first part. Will have to prove it. $\endgroup$ – AlgebraicGeometryStudent Oct 6 '13 at 1:42
  • $\begingroup$ @EnjoysMath Check the definition of "absolutely summable" again! $\endgroup$ – Pedro Tamaroff Oct 6 '13 at 1:44
  • $\begingroup$ Why can we pick M large enough so that all the terms in Tm are some of the terms in Sn? @PedroTamaroff $\endgroup$ – FreeZe Apr 22 at 10:05
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    $\begingroup$ @Waizman $T_M$ is finite, so if you go far enough into $S_N$ you will find all these terms. $\endgroup$ – Pedro Tamaroff Apr 23 at 16:55

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