We address here a MathOverflow query, given here. (Since clicking the link produces “404 Page Not Found”, I’ll reproduce the post here.)
I have a question regarding whether or not $\pi$ encodes the sequence of primer numbers. It is common knowledge that
$\zeta (2) = \sum_{i = 1}^{\infty} \frac{1}{n^2} = \prod_{p \in \mathbb{P}} \frac{1}{1-p^{-2}} = \frac{\pi^2}{6}.$Now let $M$ denote the set of all ascending sequences of natural numbers $\geq 2$, so, for instance, the complete sequence $(2,3,4,5,6,\ldots)$ is an element of $M$, just as the sequence of prime numbers $(2,3,5,7,\ldots)$. Now consider the function $f: M \rightarrow \mathbb{R}$, given by
$(a_1,a_2,\ldots) \mapsto \prod_{i \geq 1}\frac{1}{1-a_i^{-2}}.$So we know that for the sequence of primes, the value of $f$ is $\pi^2/6$. My question is now: Is the sequence of primes the only element of $M$ for which the value of $f$ is $\pi^ 2/6$? Or in other words: Is $f^{-1}(\pi^2/6) = \{ (2,3,5,7,\ldots) \}$?
As Gerald Edgar said below the query: “of course not”. But let us give a more constructive answer (at least more constructive in a technical sense). We describe first a very general result on sequences and series, after which the answer to the query falls out as a special case.
Let $f: \mathbb{N} \to \mathbb{R}_{\gt 0}$ be a function that decreases to $0$, and suppose that the following conditions are satisfied:
For some finite set $F \subset \mathbb{N}$, $\sum_{j \in F} f(j) \lt a$,
$\sum_{j \in F} f(j) + \left( \sum_{n \gt \max F} f(n) \right) \gt a$,
For every $k \gt \max F$, we have $f(k) \lt \sum_{n \gt k} f(n)$.
Define by recursion an increasing infinite sequence $n_j$ where $F = \{n_1, \ldots, n_i\}$ and for all $k \geq i$, we have that $n_{k+1}$ is the least integer greater than $n_k$ such that $f(n_1) + \ldots + f(n_{k+1}) \lt a$. (Note that $n_{k+1}$ exists at each step since the $f(n)$ decrease to $0$.) In other words, if we put $S = \{n_j\}_{j \in \mathbb{N}}$, then by construction:
For $k \gt \max F$, we have $k \in S$ if and only if $f(k) + \sum_{j \in S: j \lt k} f(j) \lt a$.
$\sum_{j \in S} f(j) = a.$
We first note that $\neg S \coloneqq \mathbb{N} \setminus S$ is infinite. To see this, observe that by condition 2, $\neg S$ has an element greater than $\max F$. If $\neg S$ were finite, then there exists a maximal $k \in \neg S$, so that $S = \{j \in S: j \lt k\} \cup \{n: n \gt k\}$. Then we have
$\sum_{j \in S} f(s) \leq a$ (since for all finite partial sums we have $\sum_{j \in S: j \lt n} f(j) \lt a$);
therefore $(\sum_{j \in S: j \lt k} f(s)) + (\sum_{n \gt k} f(n)) \leq a$;
hence $(\sum_{j \in S: j \lt k} f(s)) + f(k) \lt a$ (by condition 3);
hence $k \in S$ by the lemma, contradicting $k \in \neg S$.
Now we show $\sum_{j \in S} f(j) = a$. Suppose instead that $\sum_{j \in S} f(j) \lt a$. Pick $k \in \neg S$ so large that $f(k) + \sum_{j \in S} f(j) \lt a$. Then also $f(k) + \sum_{j \in S: j \lt k} f(j) \lt a$, but this implies $k \in S$ by the lemma, contradiction.
There are multiple infinite sets $S$ of integers greater than $1$ such that
One such set is the set of primes. But there are many such sets. For example, putting $n_1 = 2$, $n_2 = 3$, and $n_3 = 4$, we have
and then, proceeding recursively, for $k \geq 3$ we define $n_{k+1}$ to be the least integer greater than $n_1, \ldots, n_k$ such that
This places us essentially in the situation of the preceding section, by putting $f(n) = \log \frac1{1-n^{-2}}$ and $a = \log \frac{\pi^2}{6}$. All we need to check is condition 3, which here boils down to the statement that for $k \gt 4$ we have
which follows from the elementary inequality that
for $k \gt 4$.