any real number between $0$ and $1$ is expressible as an (infinite) sum of particular fractions If $0 < x \le 1$ then there is one and only one sequence of positive integers $(k_v)$, with$$1 < k_1 \le k_2\le k_3\le\cdots,$$for which $$x={1\over k_1}+{1\over k_1k_2}+\cdots+{1\over k_1k_2\cdots k_n}+\cdots$$
$x$ is rational if and only if the $k_v$ are all equal after some index $v_0$.
As far as the last statement is concerned, the if part is easy. I don't know how to build such a sequence. If I knew how to generate the $k_v$ I could probably show also the only if part. Hints?
I found that for $x \ne 1$, $\lfloor{xk_1}\rfloor = 1$ and also $\lfloor{k_2(xk_1-1)\rfloor} = 1$ and so on, so if I can determine $k_1$ from the first equation, I can determine also $k_2$ from the second, provided $(xk_1-1)$ kind of satisfy the same conditions as $x$. I don't know if this pattern is correct however, for I need to establish uniqueness. Also it doesn't seem so obvious at first glance how to prove that such a sequence gives $x$.
Obviously I can't determine $k_1$ from just the first equation, since for example, if $x = 1/2$ then both $k_1 = 2$ and $k_1 = 3$ can be chosen.
Also for $x = 1$ one has $x = {1\over2} + {1\over2\cdot2}+\cdots$ and that's why $\lfloor k_1x\rfloor \ne 1$
Uniqueness is easy as well to prove. For let there exist two such sequences $a_n$ and $b_n$, and let $m \ge 1$ be the first integer such that w.l.o.g. $b_m > a_m$. Then $0 = ({1\over a_m}+{1\over a_ma_{m+1}}+\cdots)-({1\over b_m}+{1\over b_mb_{m+1}}+\cdots)$. We see that the first sum is strictly greater than ${1\over a_m}$, whereas the second sum is $\le {1\over a_m}$, because from $b_m \ge a_m+1$ we obtain that the $n$-th term for $n \ge 1$ is $\le \left({1\over 1+a_m}\right)^n$. Thus the difference cannot be $0$.
 A: The key to this problem is the fact that the half-open intervals $({1\over k}, {1\over {k-1}}]$ for integer $k > 1$ partition the half-open interval $(0, 1]$.
Given $x \in (0, 1]$, put $x_0 = x$ and $k_0 = 2$ (so that $x_0 \le {1\over {k_0-1}}$). I claim that we can find a unique sequence of integers $k_n$ with $1 < k_1 \le k_2 \le \ldots$ such that the following equations and inequalities hold for some real numbers $x_n$ :
$$
\begin{align*}
x_0 = {1 \over k_1} + {x_1 \over k_1} \quad& 0 < x_1 \le {1 \over {k_1 -1}} \\
&\vdots\\
x_{n-1} = {1 \over k_n} + {x_n \over k_n} \quad& 0 < x_n \le {1 \over {k_n -1}}\\
&\vdots
\end{align*}
$$
To see this, at each stage take $k_n$ to be the unique integer such that ${1 \over {k_n}} < x_{n-1} \le {1 \over {k_n - 1}}$. Then $x_{n-1} = {1 \over {k_n}} + \epsilon_n$ where $0 < \epsilon_n \le {1 \over {k_n - 1}} - {1\over k_n} = {1 \over {k_n(k_n-1})}$. Now if we put $x_n = k_n \epsilon_n$, $k_n$ and $x_n$ will meet the requirements ($k_n  \ge k_{n-1}$, because $x_{n-1} \le {1 \over {k_{n-1}-1}}$).
Clearly we now have:
$$
x= x_0 = {1\over k_1}+{1\over k_1k_2}+\cdots+{1\over k_1k_2\cdots k_n} +  {x_n \over k_1k_2\cdots k_n}
$$
whence, given our bounds on the $k_n$ and $x_n$, we have that the r.h.s of the following equation converges to the l.h.s.:
$$
x = {1\over k_1}+{1\over k_1k_2}+\cdots+{1\over k_1k_2\cdots k_{n}} +  \ldots
$$
That gives existence of the above series expansion for $x$. For uniqueness, note that if we apply our construction to ${1\over l_1}+{1\over l_1l_2}+\cdots+{1\over l_1l_2\cdots l_{n}} +  \ldots$, where $1 < l_1 \le l_2 \ldots$, then we will find $k_1 = l_1, k_2 = l_2, \ldots$.
A: I will complete the proof by showing that if $x$ is rational, then the $k_v$ are equal after some index $v_0$. 
Let $x = x_0 = {m\over n}$. We have $x = {1\over k_1} + {x_1\over k_1}$, or $x_1 = {mk_1-n\over n}$. We see that $mk_1 - n$ is less than or equal to $m$, since by construction $x_0 = {m\over n} \le {1\over k_1-1}$. Thus if $x_1 = x_0$ the sequence will obviously repeat since the same integer $k_1$ chosen such that ${1\over k_1} < x_0 \le {1\over k_1-1}$ will satisfy ${1\over k_1} < x_1 \le {1\over k_1-1}$, so $k_2 = k_1$ and so on. Otherwise we have for some integer $p < m, x_1 = {p\over n}$. The same reasoning as above applies in this case. So either $x_1 = x_2$ or $x_2 < x_1$. Continuing this way, if no two successive $x_i, x_{i+1}$ are equal, eventually we will reach $x_k = {1\over n}$ (since each $x_i > 0$) which can be uniquely written as $\frac{1}{n+1} + \frac{1}{(n+1)^2} + \cdots$.
