On the "hydra set" $S=\left\{\sum_{n\geq 1}\frac{1}{x_n}:(x_n)_{n\geq 1}\text{ is an increasing sequence of positive natural numbers}\right\}$

Question

What are all the elements of set $$ S=\left\{ \sum_{n\geq 1}\frac{1}{x_n} \right\}$$ where $ (x_n)_{n\geq 1} $ is any increasing sequence of positive natural numbers for which $ \sum_{n\geq 1}\frac{1}{x_n} $ converges?

Answer 1

Any real number in $(0,1]$ can be written as $$\sum_{k=1}^{\infty}\frac{d_k}{2^k}$$ with $d_k\in\{0,1\}$ with infinite digits $d_k=1$. Now note that
$$\mathbb{R}^+=(0,1]\cup\bigcup_{n=1}^{\infty}(S_n,S_n+1]$$ where $S_n=\sum_{k=1}^n\frac{1}{2k-1}$ (recall that $S_n\to+\infty$, as $n\to\infty$). Hence, after rearranging the odd numbers and the powers of $2$, in ascending order, we may conclude that $$\left\{\sum_{n\geq 1}\frac{1}{x_n}:(x_n)_{n\geq 1}\text{ is a strictly increasing sequence in $\mathbb{N}^+$}\right\}=\mathbb{R}^+.$$

Answer 2

Every element of $\mathbb{Q}^+$ belongs to such set by a well-known statement about Egyptian fractions.
For instance $$ \frac{13}{19}=\frac{1}{2}+\frac{1}{6}+\frac{1}{57} = \frac{1}{2}+\frac{1}{6}+\frac{1}{114}+\frac{1}{228}+\frac{1}{456}+\ldots $$

On the other hand, if $\alpha\in\mathbb{R}^+\setminus\mathbb{Q}$ it is simple to construct a series of the wanted form which is convergent to $\alpha$. We take $k\in\mathbb{N}$ as the greater value such that $H_k<\alpha$. Then we take $n_1$ as the least integer such that $n_1\geq k+1$ and $\frac{1}{n_1}<\alpha-H_k$. Then we take $n_2$ as the least integer such that $n_2\geq n_1+1$ and $\frac{1}{n_2}<\alpha-H_k-\frac{1}{n_1}$. Continuing this way, $$ \frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{k}+\frac{1}{n_1}+\frac{1}{n_2}+\ldots = \alpha. $$ It follows that the wanted set is just $\mathbb{R}^+$.

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$$\pi = H_{12}+\frac{1}{27}+\frac{1}{744}+\frac{1}{1173268}+\ldots $$

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An interesting remark is that $(S,+)$ is a semigroup due to the hydra paradox. If we consider $\alpha,\beta\in S$ and simply add the associated representation terms of the form $\frac{2}{m}$ might appear. On the other hand the first term with such a structure (or an even larger numerator) can be written as the sum of unit fractions with larger denominators, and by iterating this process "from the left to the right" we eventually$^{(*)}$ reach a representation of $\alpha+\beta$ as the sum of unit fractions with distinct denominators. The Erdos-Straus conjecture states that the hydra we are fighting against has a "degree of ramification" bounded by $4$.

^{(*) Please see the exchange between I and Wojowu in the comments below, too.
I am really grateful to him for raising some interesting points.}

Answer 3

It might be worthwhile to point out that if $(a_n)$ is a sequence of positive terms, with $a_n\to 0$ and $\sum a_n = \infty,$ then for any $x>0,$ there exists an infinite subset $A\subset \mathbb N$ such that

$$\tag 1 x = \sum_{n\in A}a_n.$$

I'll omit the proof except to say it proceeds by finding "disjoint" subsums $S_1,S_2,\dots $ of $\sum a_n$ such that $S_1 + S_2 + \cdots = x.$ (The set $A$ is then the set of all indices used in any of the the $S_n$'s.)

It follows that if $n_1,n_2, \dots $ is any strictly increasing sequence of positive integers such that

$$\tag 2 \sum_{k=1}^{\infty}\frac{1}{n_{k}} = \infty,$$

then given $x>0,$ there's an infinite subsum of $(2)$ that converges to $x.$

Corollary: any $x>0$ is the infinite sum of reciprocals of distinct primes.