# Simultaneous Riemann Rearrangement

Here all functions are $$\mathbb R \to \mathbb R$$.

Fix $$M$$ a positive integer. For $$i = 0, 1, ..., M,$$ let $$f_0 = Id$$, and the other $$f_i$$ be continuous functions such that for all $$0 \leq k < M$$, $$f_{k+1}$$ is $$o(f_k)$$ as their argument goes to $$0$$.

Suppose $$x_k$$ is a sequence of reals such that $$\sum_{k=0}^{\infty} f_i (x_k)$$ is conditionally but not absolutely convergent for all $$i$$. Is it true that given any $$M+1$$ real numbers $$n_i$$, there exists a bijection $$s: \mathbb N \to \mathbb N$$ such that for each i, $$\sum_{k=0}^{\infty} f_i (x_{s(k)})$$ converges to $$n_i$$?

• I think the answer is no, but I don't have a counterexample, so I'm not 100% sure. The Lévy-Steinitz theorem asserts that the set of sums of rearrangements of the series of vectors $v_k = (f_0(x_k),\dotsc,f_M(x_k))$ is an affine subspace of $\mathbb{R}^{M+1}$, and not necessarily the whole space. E.g., the vectors $v_k = ((-1)^k/k, 1/2^k-(-1)^k/k)$ (for $k \geq 1$) have rearrangement-sums $(\alpha,1-\alpha)$, the line $x+y=1$. This example fails the requirement $f_{i+1}=o(f_i)$. I suspect that doesn't make any difference, but...? – Zach Teitler Feb 21 at 8:45

Yes -- no matter what real numbers $$n_i$$ you choose, there is a bijection $$s: \mathbb N \rightarrow \mathbb N$$ such that $$\sum_{k=0}^\infty f_i(x_{s(k)}) = n_i$$ for each $$i$$.
This follows from one form of the Lévy-Steinitz rearrangement theorem. This theorem is often quoted (as in the Wikipedia article I linked to) as saying that "the set of all sums of rearrangments of a given series of vectors in a finite-dimensional real Euclidean space is either the empty set or a translate of a subspace (i.e., a set of the form $$v + L$$, where $$v$$ is a given vector and $$L$$ is a linear subspace)."
But a stronger version of the theorem -- the version actually proved in the original papers of Lévy and Steinitz -- tells you not only that the set of all rearranged sums has the form $$v+L$$, but it tells you what $$L$$ is.
Specifically, let $$K$$ denote the set of all $$v \in \mathbb R^{M+1}$$ such that your series becomes absolutely convergent when projected onto $$v$$. Then $$K$$ is a subspace of $$\mathbb R^{M+1}$$, and $$L$$ is its orthogonal complement.
For example, if $$a_n = \frac{(-1)^n}{n}$$ and $$b_n = \frac{(-1)^n}{3n}$$, then the vector sum $$\sum_{n = 1}^\infty \langle a_n,b_n \rangle$$ becomes absolutely convergent when you project it onto $$\langle -1,3 \rangle$$. Both components of the sum are conditionally convergent, but there is a "bad direction" in which they conspire to become absolutely convergent. The Lévy-Steinitz theorem says that you can rearrange the sum to get anything you like, as long as the difference with the original sum is orthogonal to this bad direction.
In your problem, the condition you impose on the $$f_i$$ implies that any given linear combination of the $$f_i(x_k)$$ will (asymptotically) be equal to some multiple of some $$f_i$$ (whichever has the smallest index). Thus there is no "bad direction" for your sum -- it remains conditionally convergent no matter what vector in $$\mathbb R^{M+1}$$ you project it onto. By the Lévy-Steinitz theorem, it follows that any vector in $$\mathbb R^{M+1}$$ can be obtained as the sum of some rearrangement of your series.