An explicit greedy rearrangement of the alternating harmonic series with sum $0$, and the same recipe for any prescribed real

Let $a_j=(-1{)}^j\mathrm{/}\iota (j+1)$ be the terms of the alternating harmonic series, which converges conditionally (${\sum }_{j\ge 0}(-1{)}^j\mathrm{/}(j+1)$ converges conditionally, with sum strictly between $1\mathrm{/}2$ and $1$). Fix a real $c$. The greedy rearrangement towards $c$ is the bijection $\sigma$ of $\mathbb{N}$ produced by the following rule, which is exactly the construction of The Riemann series theorem: a conditionally convergent real series has, for every $c\in \mathbb{R}$, a rearrangement with sum $c$, and rearrangements diverging to $+\mathrm{\infty }$, to $-\mathrm{\infty }$, and oscillating with any prescribed $\mathrm{lim\hspace{0.17em}inf}\le \mathrm{lim\hspace{0.17em}sup}$ in $\bar{\mathbb{R}}$ with the constant target $c$:

at each step, if the running sum of the terms already used is at most $c$, take the next unused nonnegative term of the series; otherwise take the next unused negative term.

By The Riemann series theorem: a conditionally convergent real series has, for every $c\in \mathbb{R}$, a rearrangement with sum $c$, and rearrangements diverging to $+\mathrm{\infty }$, to $-\mathrm{\infty }$, and oscillating with any prescribed $\mathrm{lim\hspace{0.17em}inf}\le \mathrm{lim\hspace{0.17em}sup}$ in $\bar{\mathbb{R}}$ the resulting rearrangement converges, with

${\sum }_{k=0}^{\mathrm{\infty }}{a}_{\sigma (k)}=c\mathrm{.}$

For $c=0$ the rule produces, in order,

$1,-\frac{1}{2},-\frac{1}{4},-\frac{1}{6},-\frac{1}{8},\frac{1}{3},-\frac{1}{10},-\frac{1}{12},-\frac{1}{14},-\frac{1}{16},\frac{1}{5},\dots$

the running sums after the successive terms being $1,\frac{1}{2},\frac{1}{4},\frac{1}{12},-\frac{1}{24}$, then $\frac{7}{24}$ after $\frac{1}{3}$, and so on: one positive term followed by however many negative terms are needed to bring the running sum below $0$ again.

The same series therefore has rearrangements summing to $0$, to $S$ itself, to $\frac{3}{2}S$ (Taking two positive terms for each negative one rearranges the alternating harmonic series to $3\mathrm{/}2$ times its sum, by the identity $T_{3n}=S_{4n}+\frac{1}{2}{S}_{2n}$) and to every other real number, while its terms are never changed.

• The displayed initial segment is arithmetic, not a further claim. Starting from a running sum of $0$, which is at most $c=0$, the rule takes the first nonnegative term $1$; the running sum $1$ then exceeds $0$, so negative terms are taken until it drops below $0$, which the successive values $\frac{1}{2},\frac{1}{4},\frac{1}{12},-\frac{1}{24}$ show happens after four of them. The reader may continue the computation indefinitely; nothing in the verification depends on it.

• Why the greedy rule terminates each phase. Each phase ends because the relevant part series diverges to $+\mathrm{\infty }$ (Positive and negative parts: $a_k=a_k^{+}-a_k^{-}$ and $\mathrm{\mid }{a}_k\mathrm{\mid }=a_k^{+}+a_k^{-}$; a series converges absolutely iff both $\sum a_k^{+}$ and $\sum a_k^{-}$ converge, and for a conditionally convergent series both diverge to $+\mathrm{\infty }$), so however far along the supply one is, enough remains to cross the target. This is the only place conditional convergence is used, and it is the reason the example has no analogue for an absolutely convergent series.

• The overshoot vanishes, which is why the sum is exactly $c$. At each crossing the running sum differs from $c$ by at most the term just used, and the terms of a convergent series tend to $0$; since both supplies are consumed in order and both are exhausted, the terms used at successive crossings tend to $0$ as well.