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I know that any conditional convergent series has a rearrangement that diverges. For example if we have $$ \sum_{n=1}^\infty \frac{(-1)^n}{ n} $$ what is a rearrangement that diverges?

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Add "positive terms" until you exceed $1$. Then add the first "negative term". Add positive terms until you exceed $2$. Add the second negative term. Add ... –  David Mitra Aug 24 at 13:13
    
@DavidMitra Can this rearrangement be written in explicit form? –  user150391 Aug 24 at 14:23
    
It might be nasty to do so for what I wrote. But, take something like $S_1-1+S_2-{1\over2}+S_3-{1\over3}+\cdots$ where $S_1={1\over2}+{1\over4} $, $S_2={1\over 6}+{1\over 8}+{1\over 10}+{1\over12}$, $S_3={1\over14}+\cdots+{1\over28}$, $\ldots$. Each block $S_n$ is larger than $1/4$ (if I calculated things correctly). The series diverges since its partial sums are not Cauchy. –  David Mitra Aug 24 at 15:18
    
@DavidMitra If you write your comments as an answer I will accept and upvote it. –  user150391 Aug 24 at 15:18

2 Answers 2

up vote 3 down vote accepted

Let $S=\sum\limits_{n=1}^\infty (-1)^n a_n$, $a_n\ge0$, be conditionally convergent.

To obtain a divergent rearrangement of $S$, you can take advantage of the fact that the sum of the positive terms of $S$ diverges to $\infty$. Then, given any $N$, you can find an $M $ so that the sum of successive positive terms, $a_{2N}+a_{2(N+1)}+\cdots+ a_{2( N+M)}$ is as large as you like.

You then construct the rearrangement by taking the sum of successive positive terms until you exceed $1$ (or any fixed positive number). Then add the first negative term. Then add the next block of successive positive terms whose sum exceeds $1$. Then add the next negative term. And so on.

You need to be careful to exhaust all terms of the series (so that you indeed have a rearrangement). If you do, then the resulting rearrangement will diverge as the resulting sequence of partial sums will not be Cauchy.

One can write an explicit example of a divergent rearrangement (as I attempted to do in the comments for your example).

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Not sure if it helps but:

$$\sum_{n=1}^\infty \frac{(-1)^n}{n} = \underbrace{\sum_{n=1}^\infty \frac{1}{2n}}_{S_1} - \underbrace{\sum_{n=1}^\infty \frac{1}{2n-1}}_{S_2}$$

But $S_1$ and $S_2$ diverge.

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