Let $\mathbb D$ be the set of all real numbers that can be expressed as a sum of distinct prime reciprocals, i.e. $\mathbb D = \{ d \in \mathbb R \mid d = \sum_{k \in \mathbb K} \frac 1k $ for some (possibly infinite) subset $\mathbb K$ of the prime numbers $\mathbb K \subset \mathbb P \}$.

I'm not a mathematician, and I have only a very vague idea of infinite sums, but intuitively, it seems to me that most real numbers are not in $\mathbb D$. Is that true? Is there any way to decide whether a given real number is in $\mathbb D$, e.g. is $\pi \in \mathbb D$?


EDIT, next morning: when I wrote the first paragraph below, it seemed obvious to me that the greedy algorithm worked, based on minimal hypotheses. I have since become uncertain about that, and now suspect that Bertrand's is crucial in what I wrote below... I asked a question to follow up, and am completely satisfied with one answer. i remain uncertain about the question of whether the greedy algorithm always works; that is not what I asked in my question. In the other direction, i am having no luck contriving a sequence (of decreasing reals0 and a positive target for which greedy fails, so... see Can every positive real be written as the sum of a subsequence of dot dot dot

ORIGINAL: All positive real numbers are in $\mathbb D.$ The sum of prime reciprocals diverges but the individual terms go to zero. A greedy algorithm for choosing primes to be included will give anything desired; that is, given whatever is left after subtracting off the prime reciprocals used so far, take the smallest prime with reciprocal smaller than what remains.

I can be a bit more specific about the progress at each step, because of Bertrand's Postulate. Call the sum so far $R,$ and the target $T > R.$ We finally get to some point where a choice arises, because the reciprocal of the very next prime is too large. As the primes grow, eventually there are consecutive primes $p,q$ such that $p$ does not work but $q$ does. That is, $$ R + \frac{1}{p} > T \geq R + \frac{1}{q}.$$ Subtract $R$ throughout, $$ \frac{1}{p} > T -R \geq \frac{1}{q}.$$ Invert, $$ p < \frac{1}{T-R} \leq q. $$ Bertrand's Postulate says that $q < 2p,$ so we have $$ p < \frac{1}{T-R} \leq q < 2 p < \frac{2}{T-R}. $$ Invert back, $$ \frac{1}{p} > T -R \geq \frac{1}{q} > \frac{1}{2 p} > \frac{T-R}{2}.$$ Add $R$ throughout, $$ R + \frac{1}{p} > T \geq R + \frac{1}{q} > R + \frac{1}{2 p} > \frac{T+R}{2},$$ or $$ T \geq R + \frac{1}{q} > \frac{T+R}{2}.$$ That is, once we get close enough, we can do at least as well as halving the remaining distance to $T.$

Put the diverging series and the halving thing together, and the process can be summarized in a hybrid manner, that says we can halve the distance to go at each "step." We have already used some primes and have a distance still to go. If the reciprocal of the next prime is small enough, and the next prime after that, and so on. Eventually we arrive at the choice described above, and the distance to go is smaller. By one final choice, above $q$ rather than $p,$ we can halve what remains. So, call that one step, a string of consecutive primes (possibly empty) followed by a singleton choice.

This method leaves me wondering, given any strictly decreasing sequence of positive numbers with divergent sum, but no Bertrand's, can we still say we get all positive reals? I am not sure.

  • $\begingroup$ So that would be a general principle: If xk > 0, limit of xk -> 0, and sum of xk -> infinity, then every number > 0 is the limit of sum sub series of xk. $\endgroup$
    – gnasher729
    Jun 20 '14 at 9:54
  • $\begingroup$ Do I understand correctly that the greedy choice is not strictly necessary? For example, can you leave out all primes below 1000 and still converge to any positive real number? $\endgroup$
    – gatinueta
    Jun 20 '14 at 11:15
  • 1
    $\begingroup$ @gatinueta, apparently so. I asked a follow-up question and am satisfied with this answer: math.stackexchange.com/questions/840286/… I am still uncertain about whether the greedy choice works for every sequence, but i do not believe i will ask a third question on this. the greedy algorithm does work for prime reciprocals. $\endgroup$
    – Will Jagy
    Jun 20 '14 at 16:39

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