Show that the sum of an absolutely convergent series does not change if the terms are rearranged.

Let the absolutely convergent series be $a_1,a_2,\ldots$, and let the rearrangement be $b_1,b_2,\ldots$. By the Cauchy criterion of convergence, since the series $|a_1|,|a_2|,\ldots$ converges, it follows that for any $\epsilon$, there exists $N$ such that $|a_m|+|a_{m+1}|+\ldots+|a_n|<\epsilon$ for all $n>m>N$.

We'll show that the series $b_i$ converges. Let $N'$ be the greatest index such that $b_{N'}$ comes from one of $a_1,a_2,\ldots,a_N$. Then for all $n>m>N'$, we have $|b_m+b_{m+1}+\ldots+b_n|\leq |b_m|+|b_{m+1}|+\ldots+|b_n|<\epsilon$.

So the series $b_i$ converges. How can we show it converges to the same sum as $a_i$?

  • $\begingroup$ You covered it well. For the rest, assume they converge to two distinct limits $0\leq c_1<c_2$. Find $0<\epsilon<c_2-c_1$, and the rest should follow. $\endgroup$ – Jonathan Y. Aug 25 '13 at 14:30
  • $\begingroup$ @JonathanY. And what do we do with that $\epsilon$? I still can't see it. $\endgroup$ – Mika H. Aug 25 '13 at 14:50

Say w.l.o.g. $\sum a_n=c_1<c_2=\sum b_n$. Take $0<\epsilon <\frac{c_2-c_1}{3}$. Then take $N>0$ such that for all $m>N$ we have $$\left|\sum_{n\leq m} b_n - c_2\right|<\epsilon; \quad \sum_{n>N}|b_n|<\epsilon$$ and take $M>0$ such that $\{a_1,\ldots,a_M\}$ contain $\{b_1,\ldots,b_N\}$ and $$\left|\sum_{n\leq M} a_n - c_1\right|<\epsilon.$$ Then $$c_2-c_1< (\sum_{n\leq N} b_n + \epsilon) - (\sum_{n\leq M} a_n - \epsilon)\leq 2\epsilon + \sum_{n>N}|b_n| < 3\epsilon< c_2-c_1.$$


A) First consider a nonnegative-term series $\displaystyle\sum_{n=1}^{\infty} a_n$, and let $\displaystyle\sum_{n=1}^{\infty} b_n$ be any rearrangement of $\displaystyle\sum_{n=1}^{\infty} a_n$.
If $(S_n)$ and $(T_n)$ are the sequences of partial sums for $\displaystyle\sum_{n=1}^{\infty} a_n$ and $\displaystyle\sum_{n=1}^{\infty} b_n$, then $T_n\le S_N$ for each $n$, where $\displaystyle N=\max_{1\le i\le n}\{k:b_{i}=a_{k}\}$, so $\displaystyle\sum_{n=1}^{\infty} b_n\le \sum_{n=1}^{\infty} a_n$.

Since $\displaystyle\sum_{n=1}^{\infty} a_n$ is also a rearrangement of $\displaystyle\sum_{n=1}^{\infty} b_n$,

we also have that $\displaystyle\sum_{n=1}^{\infty} a_n\le \sum_{n=1}^{\infty} b_n$; $\;\;\;$so $\displaystyle\sum_{n=1}^{\infty} b_n= \sum_{n=1}^{\infty} a_n$.

B) In the general case, let $\displaystyle\sum_{n=1}^{\infty} a_n=\sum_{n=1}^{\infty} a_n^{+}-\sum_{n=1}^{\infty} a_n^{-}$ and $\displaystyle\sum_{n=1}^{\infty} b_n=\sum_{n=1}^{\infty} b_n^{+}-\sum_{n=1}^{\infty} b_n^{-}$

where $a_{n}^{+}=\frac{a_{n}+|a_{n}|}{2}$ and $a_{n}^{-}=\frac{|a_{n}|-a_{n}}{2}$ and similarly for $b_{n}^{+}$ and $b_{n}^{-}$.

Since $\displaystyle\sum_{n=1}^{\infty} |a_n|=\sum_{n=1}^{\infty} a_n^{+}+\sum_{n=1}^{\infty} a_n^{-}$ and $\displaystyle\sum_{n=1}^{\infty} |b_n|=\sum_{n=1}^{\infty} b_n^{+}+\sum_{n=1}^{\infty} b_n^{-}$,

from Part A) we get that

$\;\;\;\;\;\;\;\;\;\displaystyle\sum_{n=1}^{\infty} b_{n}^{+}=\sum_{n=1}^{\infty} a_n^{+}$ and $\displaystyle\sum_{n=1}^{\infty} b_{n}^{-}=\sum_{n=1}^{\infty} a_n^{-}$;

so $\displaystyle\sum_{n=1}^{\infty} b_n=\sum_{n=1}^{\infty} b_n^{+}-\sum_{n=1}^{\infty} b_n^{-}=\sum_{n=1}^{\infty} a_n^{+}-\sum_{n=1}^{\infty} a_n^{-}=\sum_{n=1}^{\infty} a_n$.


Apply reasoning to $\sum (a_i - b_i)$. For given $\epsilon > 0$, let $N$ be the index in such that partial sums of $|a_i|$ after index $N$ are less than $\epsilon / 2$. Then choose $N'$ so that all of $a_1,\ldots,a_N$ appear as terms in $b_1,\ldots,b_{N'}$. Then consider the partial sum of $(a_i - b_i)$ up to any index $M > N'$. Rearrange the ordering of the first $N'$ terms in $b_i$ so that the first $N$ terms are the first $N$ terms of $a_i$ so that the first $N$ terms of $(a_i - b_i)$ are all $0$. Then note that the absolute value of the sum of remaining terms $\sum_{i=N+1}^{M} (a_i - b_i)$ is less than or equal to $\sum_{i=N+1}^{M} |a_i|$ + $\sum_{i=N+1}^M |b_i| < \epsilon / 2 + \epsilon / 2 = \epsilon$.

  • $\begingroup$ Why is it that $\sum_{i=N+1}^M|b_i|<\epsilon/2$? $\endgroup$ – Mika H. Aug 25 '13 at 15:19

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