TITLE: An algorithm for making conditionally convergent series take arbitrary values?
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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
$$T=\sum_{n=0}^\infty \left(\sum_{i=p_n+1}^{p_{n+1}} \frac{1}{i} - \sum_{j=q_n+1}^{q_{n+1}} \frac{1}{j}\right)$$