Does this infinite primes snake-product converge?

Solution 1:

A very extended comment explaining why this problem is probably difficult.

Let $g_n=p_{2n}-p_{2n-1}$. The product we are looking at is then $$\prod_{n=1}^\infty\left(1-\frac{g_n}{p_{2n}}\right)^{(-1)^n}.$$ Taking logarithms, we are met with a sum of the form $$\sum_{n=1}^\infty\left((-1)^{n-1}\frac{g_n}{p_{2n}}+O\left(\left(\frac{g_n}{p_{2n}}\right)^2\right)\right).$$ Using results due to Heath-Brown on second moments on prime gaps (see here), namely $\sum_{k=1}^ng_k^2=O(x^{7/6+\varepsilon})$, by summation by parts we can bound the sum of the error terms by a finite value.

Hence we are left with an alternating sum of $g_n/p_{2n}$. To deal with this, we essentially have to show the sums $\sum_{n=1}^N(-1)^{n-1}g_n$ are asymptotically smaller than $p_{2N}$ (this won't guarantee convergence, but we definitely want that to hold). Using the notation of this MO answer (and the paper it cites), this sum is equal to $S(2N;1,4)-S(2N;3,4)$. What we would like to know is that this difference is $o(p_{2N})$. So you see we are quickly lead to investigating asymptotics of $S(N;a,q)$. Conjecturally, we have $$S(N;a,q)\sim\frac{p_N}{q}$$ (so that the gaps are in some sense equidistributed), but available bounds are much weaker. In the cases we are interested in, we only get $$\liminf\frac{S(N;a,4)}{p_N}\geq\frac{1}{256}$$ unconditionally, and even conditionally on prime tuples conjecture we get $\geq 1/32$, while what we would like is for the limits to exist and be equal.

Hence, as you can see, the available methods are not capable of showing that the difference $S(2N;1,4)-S(2N;3,4)$ is asymptotically small.