# An algorithm for making conditionally convergent series take arbitrary values?

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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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

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.

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
$$T=\sum_{n=0}^\infty \left(\sum_{i=p_n+1}^{p_{n+1}} \frac{1}{i} – \sum_{j=q_n+1}^{q_{n+1}} \frac{1}{j}\right)$$