A question about alternate series involving unit fractions I don't know exactly how to classify this question. It is not from any homeworks, just something I've been wondering about.
Let $A\subseteq\mathbb N$ be a subset that contains at least $n$ elements; then $A(n)$ is the $n$-th element of $A$, and 
let $R$ be a function $R:P(\mathbb N) \rightarrow [0,1]$.
If $A$ is not empty and the cardinality of $A$ is $|A|$, then $$R(A)=\sum_{n=1}^{|A|}\frac{(-1)^{n+1}}{A(n)} $$
and $R(\emptyset )=0$.
 Pierce expansion shows that $R$ is onto, and I know that $R$ isn't one-to-one.
My question is whether for every real number $r$ between $0$ and $1$ the inverse image $R^{-1}[\{r\}]$ is finite. Or is it only finite for $0$, $\frac{1}{2}$ and $1$?
 A: We get an infinite preimage for any number other than 0,0.5,1.
First note that if we have some "partial series" $\sum_{k=0} ^n (-1)^{k}A(k)^{-1}$ (with A a monotone function of k into the naturals), a necessary and sufficient condition for a continuation of the series to exist which converges to $x$ (that is, some $\tilde A\in R^{-1}(x)$ s.t. $\tilde A (k)=A(k)$ for all $k\le n$,) is that there is some $B\in R^{-1}\left|((\sum_{k=0} ^n (-1)^{k}A(k)^{-1}) - x)\right|$ with $min(B)>max(A)$. Excuse the abuse of notation: A,B are functions, and to consider them as sets we look at the functions' respective images.
Now let's show $R^{-1}(x)$ is infinite for $0<x<\frac{1}{10}$.
There exists a natural $n$ with $\frac{1}{n+2}<x<\frac{1}{n}$. We have:
$\frac{1}{n}-x < \frac{1}{n}-\frac{1}{n+2}=\frac{2}{n(n+2)}< \frac{2}{10n}<\frac{1}{5n}$
$\frac{1}{n-2}-x < \frac{1}{n-2}-\frac{1}{n+2}=\frac{4}{n^2 -4}<\frac{4}{10n-n}=\frac{1}{2.25n}$
So the differences $\frac{1}{n}-x$ and $\frac{1}{n-1}-x$ are both bounded strictly above by $\frac{1}{2.2n}$. Therefore there exists some $m\ge 2n$ s.t. 
$\frac{1}{m+2}<\frac{1}{n}-x<\frac{1}{m}$ 
and some similar $\tilde{m}$ with:
$\frac{1}{\tilde m + 2}<\frac{1}{n-1}-x<\frac{1}{\tilde m}$. 
We can begin an alternating series with either $\frac{1}{n}$ or $\frac{1}{n-1}$, and at each stage concatenate to the series either of two values $(-1)^{k} m^{-1} , (-1)^k (m-1)^{-1}$,  where $m$ satisfies the previous inequality. Such an $m$ will be strictly larger than $n$ (in fact larger than $2n$). The next value appended (denote it p) will have to satisfy an inequality of the form:
$\frac{1}{p+2}<\left|\frac{1}{n}-\frac{1}{m}-x\right|<\frac{1}{p}$.
Continuing this way, at each stage we can append one of two values (of the form $(m-1)^{-1}$ or $m^{-1}$ times an appropriate sign) to the alternating series. The resulting sums both approach x (in fact the number of accurate digits is nearly doubled at each stage).
To extend this construction to $\frac{1}{3}$, note that $\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{9}\right) - \frac{1}{3}<\frac{1}{10}$. From there we can continue as already shown.
To extend the same procedure to any $x<\frac{1}{2}$ shouldn't be difficult (the calculations are similar - we proceed until at some stage we have a remainder less than $\frac{1}{10}$ from the finite alternating sum). To extend it to $\frac{1}{2}<x<1$ note $1-x<\frac{1}{2}$. I haven't done it on paper but I would be surprised to hear $R^{-1}$ is finite for any more values.
A: Certainly $R^{-1}(1/3)$ is infinite, via the following sequence:
$1/3$
$1/2 - 1/6$
$1/2 - (1/5 - 1/30)$
$1/2 - (1/5 - (1/29 - 1/870))$
Each expression is an alternating sum that adds to $1/3$, where the last term $N$ in the previous subset is replaced by $\{N-1, N(N-1)\}$. You can always repeat this unless your two largest elements of $A$ are consecutive, so that answers your question for most numbers. To clarify, are you restricting $A$ to finite subsets?
Edit: If the two largest elements of $A$ are consecutive, i.e. $N, N+1 \in A$, then replacing both by $N(N+1)$ gives the same value under $R$, so this procedure along with the above one gives infinite preimage for all rational numbers in $[0,1]$, except when $N-1$ cannot be added, or $N = N(N-1)$. These scenarios only occur when $A = \{1\}$ or $A = \{1/2\}$. For irrational numbers I have no idea.
