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This thread reminded me of an old unsettled question I have.

Given an arbitrary conditionally convergent series $\beta=\sum\limits_{k=1}^\infty a_k$ and a target value $\alpha$, is there an algorithm for finding the permutation of the original series that will make it sum to $\alpha$? Alternatively, if the permutation cannot be explicitly given, is there an algorithm for finding the first few terms of the rearrangement of the series for $\beta$ to make it sum to $\alpha$?

So far, what I've seen is a method for rearranging the alternating harmonic series $\log\,2=\sum\limits_{k=1}^\infty \frac{(-1)^k}{k}$ in Stan Wagon's Mathematica in Action. I would like to know if the method there is generalizable to arbitrary conditionally convergent series.

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    @Sri: That settles the alternative question, it looks. But let's take the $\log\,2$ series again as an example. Is there an algorithm that starts from the series and the target value $\log\sqrt 2$ and returns the series $$\sum_{k=0}^\infty \left(\frac1{2k+1}-\frac1{4k+2}-\frac1{4k+4}\right)$$?2011-09-08
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    For specific "nice" target values, there could be "nice" rearrangements that does the job. I am sure how we can tell the algorithm: "$\log \sqrt{2}$ is nice, so I prefer a nice rearrangement as my answer". (By the way, have you tried Ross's algorithm with $\log \sqrt{2}$. What rearrangement does that give?)2011-09-08
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    I'm far away from my box with *Mathematica*; so I'll try programming Ross's algorithm later. For now, I don't know what the algorithm will give for that case.2011-09-08
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    Actually, the "overshoot-undershoot" algorithm seems to work for $\alpha = \log \sqrt{2}$ to give the rearrangement you have in mind. I haven't proved it, but I tried it in a calculator for the first 5 summands ($k=0$ till $k=4$ in your new series). (By the way, a small nitpick. The way you have written the series for $\log \sqrt{2}$, it is not a rearrangement of the $\log 2$ series. In fact, the new series is even absolutely convergent :-). But I get what you mean, of course...)2011-09-08
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    @Sri: Point taken. :) I suppose the better way would be to say I'm looking for an algorithm that gives a new series expression equivalent to what you would get if you permuted the original series.2011-09-08
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    @J.M., that's a bit much to call for unless you have some serious restrictions on the target value in mind. Otherwise, there are uncountably many possible targets, but at most countably many nicely-shaped series expressions. (Yes, this argument is a little frayed at the edges, since we can arguably assume that the target can be finitely described _somehow_ or we wouldn't be able to run an algorithm on it -- but if we accept arbitrary finite descriptions of the target, then the over/undershoot algorithm shoud also _itself_ count as a "series expression".)2011-09-08

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