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I need some help on a challenge problem from my analysis class this week. The question is whether the series

$s = \displaystyle \sum_{n \in \mathbb{N}} a_n = \sum \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} \ldots...$

Can be re-arranged in such a way so as not to converge?

Chapter 3 of Principles of Mathematical Analysis by Rudin includes an example where the series can be re-arranged so as not to converge to the same value as $s$.

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    Reciprocals of odd numbers from $2^{n-1}+1$ to $2^{n}-1$, followed by negative of reciprocal of $2n$, $n=1,2,\dots$.2012-10-14

1 Answers 1

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Yes, it can. It can also be rearranged to converge to any specified real number. This is true of every conditionally convergent series.

To rearrange it to diverge to $\infty$, start by adding positive terms until the total is at least $1$; that takes just one term, $1$. Then add the first negative term that hasn’t yet been included; in this case you just get $1-\frac12$. That’s step $1$. Step $2$ is to add positive terms, starting where you left off in step $1$, until the partial sum is at least $2$, and then throw in the first unused negative term; you get

$1-\frac12+\frac13+\frac15+\ldots+\frac1{41}-\frac14\;.$

In general at step $n$ you add positive terms, starting where you left off in step $n-1$, until the partial sum is at least $n$, and then add the next negative term.

This procedure can be carried out because the series $1+\frac13+\frac15+\frac17+\ldots$ diverges, so that at step $n$ we can always add enough positive terms to bring the partial sum up to $n$. And since the rearranged series has arbitrarily large partial sums, it must diverge.

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    @Elements: Yes, it was just a mistake. You actually have to go to $\frac1{41}$, if my hasty arithmetic is correct. I thought that the description made the general method pretty clear. If you want a rearrangement that sums to $5$, say, add positive terms until you get a partial sum >5, then include the first negative term; then add more positive terms (in order) until you go over $5$ again, and throw in the next negative term; and so on. Use the conditional convergence of the series to show that this can be carried out for any desired sum, including $\infty$ and $-\infty$.2012-10-17