Todd Trimble
elementary sequence question

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 π\pi encodes the sequence of primer numbers. It is common knowledge that

ζ(2)= i=1 1n 2= p11p 2=π 26. \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 MM denote the set of all ascending sequences of natural numbers 2\geq 2, so, for instance, the complete sequence (2,3,4,5,6,)(2,3,4,5,6,\ldots) is an element of MM, just as the sequence of prime numbers (2,3,5,7,)(2,3,5,7,\ldots). Now consider the function f:Mf: M \rightarrow \mathbb{R}, given by

(a 1,a 2,) i111a i 2. (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 ff is π 2/6\pi^2/6. My question is now: Is the sequence of primes the only element of MM for which the value of ff is π 2/6\pi^ 2/6? Or in other words: Is f 1(π 2/6)={(2,3,5,7,)}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: >0f: \mathbb{N} \to \mathbb{R}_{\gt 0} be a function that decreases to 00, and suppose that the following conditions are satisfied:

  1. For some finite set FF \subset \mathbb{N}, jFf(j)<a\sum_{j \in F} f(j) \lt a,

  2. jFf(j)+( n>maxFf(n))>a\sum_{j \in F} f(j) + \left( \sum_{n \gt \max F} f(n) \right) \gt a,

  3. For every k>maxFk \gt \max F, we have f(k)< n>kf(n)f(k) \lt \sum_{n \gt k} f(n).

Define by recursion an increasing infinite sequence n jn_j where F={n 1,,n i}F = \{n_1, \ldots, n_i\} and for all kik \geq i, we have that n k+1n_{k+1} is the least integer greater than n kn_k such that f(n 1)++f(n k+1)<af(n_1) + \ldots + f(n_{k+1}) \lt a. (Note that n k+1n_{k+1} exists at each step since the f(n)f(n) decrease to 00.) In other words, if we put S={n j} jS = \{n_j\}_{j \in \mathbb{N}}, then by construction:

Lemma

For k>maxFk \gt \max F, we have kSk \in S if and only if f(k)+ jS:j<kf(j)<af(k) + \sum_{j \in S: j \lt k} f(j) \lt a.

Proposition

jSf(j)=a.\sum_{j \in S} f(j) = a.

Proof

We first note that ¬SS\neg S \coloneqq \mathbb{N} \setminus S is infinite. To see this, observe that by condition 2, ¬S\neg S has an element greater than maxF\max F. If ¬S\neg S were finite, then there exists a maximal k¬Sk \in \neg S, so that S={jS:j<k}{n:n>k}S = \{j \in S: j \lt k\} \cup \{n: n \gt k\}. Then we have

  • jSf(s)a\sum_{j \in S} f(s) \leq a (since for all finite partial sums we have jS:j<nf(j)<a\sum_{j \in S: j \lt n} f(j) \lt a);

  • therefore ( jS:j<kf(s))+( n>kf(n))a(\sum_{j \in S: j \lt k} f(s)) + (\sum_{n \gt k} f(n)) \leq a;

  • hence ( jS:j<kf(s))+f(k)<a(\sum_{j \in S: j \lt k} f(s)) + f(k) \lt a (by condition 3);

  • hence kSk \in S by the lemma, contradicting k¬Sk \in \neg S.

Now we show jSf(j)=a\sum_{j \in S} f(j) = a. Suppose instead that jSf(j)<a\sum_{j \in S} f(j) \lt a. Pick k¬Sk \in \neg S so large that f(k)+ jSf(j)<af(k) + \sum_{j \in S} f(j) \lt a. Then also f(k)+ jS:j<kf(j)<af(k) + \sum_{j \in S: j \lt k} f(j) \lt a, but this implies kSk \in S by the lemma, contradiction.

Application

There are multiple infinite sets SS of integers greater than 11 such that

jS11j 2=π 26.\prod_{j \in S} \frac1{1 - j^{-2}} = \frac{\pi^2}{6}.

One such set is the set of primes. But there are many such sets. For example, putting n 1=2n_1 = 2, n 2=3n_2 = 3, and n 3=4n_3 = 4, we have

j=1 311n j 2=1.6<π 26\prod_{j = 1}^3 \frac1{1 - n_j^{-2}} = 1.6 \lt \frac{\pi^2}{6}

and then, proceeding recursively, for k3k \geq 3 we define n k+1n_{k+1} to be the least integer greater than n 1,,n kn_1, \ldots, n_k such that

j=1 k+111n j 2<π 26.\prod_{j=1}^{k+1} \frac1{1 - n_j^{-2}} \lt \frac{\pi^2}{6}.

This places us essentially in the situation of the preceding section, by putting f(n)=log11n 2f(n) = \log \frac1{1-n^{-2}} and a=logπ 26a = \log \frac{\pi^2}{6}. All we need to check is condition 3, which here boils down to the statement that for k>4k \gt 4 we have

11k 2< n>k11n 2\frac1{1 - k^{-2}} \lt \prod_{n \gt k} \frac1{1- n^{-2}}

which follows from the elementary inequality that

11k 2<11(k+1) 211(k+2) 2\frac1{1 - k^{-2}} \lt \frac1{1 - (k+1)^{-2}} \cdot \frac1{1 - (k+2)^{-2}}

for k>4k \gt 4.

Revised on April 16, 2016 at 18:32:01 by Todd Trimble