Does this infinite primes snake-product converge? Form an infinite product of prime ratios as follows.
Start with
$$
\frac{2}{3}\cdot\frac{7}{5}=\frac{14}{15} \approx 0.93 \;.
$$
Continue alternating a fraction $< 1$ times the next fraction $>1$,
progressively through the primes:
$$
\frac{2}{3}\cdot\frac{7}{5}\cdot\frac{11}{13}\cdot\frac{19}{17}
= \frac{2926}{3315} \approx 0.88 \;,
$$
$$
\frac{2}{3}\cdot\frac{7}{5}\cdot\frac{11}{13}\cdot\frac{19}{17}\cdot\frac{23}{29}\cdot\frac{37}{31}
=\frac{2490026}{2980185} \approx 0.83 \;.
$$
Continue this process to $\infty$. One way to write the product is
$$
\xi = \prod_{1,5,9,\ldots}^\infty 
\frac{p_i}{p_{i+1}}\cdot\frac{p_{i+3}}{p_{i+2}}
$$
where $p_i$ is the $i$-th prime.
I call this the primes snake-product:

          


My questions are:


Q1. Does the product converge?
Q2. If so, to what value $\xi$ does it converge?

Up to the $1$-millionth prime ($15485863$),
the product is about $0.9056$:

          


Update (26Jan2019): @Peter has calculated out to $p_i=10^{10}$ when the product is
$\approx 0.9048$.
 A: 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.
