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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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    @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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The following paper might answer your question: C. C. Cowen, K. R. Davidson, and R. P. Kaufman, Rearranging the alternating harmonic series, American Mathematical Monthly 87 (1980) 17-19.

The useful (old) theorem here is:

THEOREM. Suppose the Alternating Harmonic Series is rearranged so that, if $p_n$ and $m_n$ are the numbers of positive and negative terms among the first $n$ terms, then $r = $

$\lim_{n \to \infty}\; p_n / m_n $

exists. Then series sums to $\log 2 + \frac12 \log r$.

So to hit a target $T$, just let the positive:negative ratio be $(1/4) e^{2T}$.

Of course, if this last expression turns out to be rational (as when $T = \log 2$), things are particularly nice.

Stan Wagon

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    @Did particular case: to hit the target $T=0$, we need$a$p/n ratio of $e^{-\pi}$. This can be explicitly achieved (without overshoot-undershoot-) through the series for $e^\pi$ in http://math.stackexchange.com/questions/1604718/rational-series-representation-of-e-pi. These series have rational terms, so allow for permuting conditionally convergent series. The first one is faster.2016-01-12
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The first few terms can be anything you want, because by the Riemann series theorem you can rearrange the rest to give the proper sum. In the proof of the theorem, you basically add enough positive terms to get greater than your target, then enough negative terms to get smaller, and so on. This gives an effective procedure and works on any conditionally convergent series, but it doesn't tell you what the 1000th term (say) of the rearranged series will be.

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    Thanks, that settles the "alternative" question. :) I'm still looking out for something that works on what I presented in the comments though.2011-09-08
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A slight formal modification of the series $ \log\left(\frac{p}{q}\right)=\sum_{i=0}^\infty \left(\sum_{j=pi+1}^{p(i+1)}\frac{1}{j}-\sum_{k=qi+1}^{q(i+1)}\frac{1}{k}\right) $ allows to compute numbers other than the logarithms of positive rationals.

Instead of positive integers $p,q$, consider integer sequences $p_n, q_n$ such that $p_0=q_0=0$ $p_{n+1}>p_n$ $q_{n+1}>q_n$ and $\lim_{n \to \infty} \frac{p_n}{q_n}=e^T$

The following rearrangement of the cancelling harmonic series converges to the target $T=\sum_{n=0}^\infty \left(\sum_{i=p_n+1}^{p_{n+1}} \frac{1}{i} - \sum_{j=q_n+1}^{q_{n+1}} \frac{1}{j}\right)$