$ \sum_{n = 1}^{\infty} ( \frac{p_{2n-1}}{p_{2n}} - \frac{p_{2n}}{p_{2n +1}} ) = ?? $

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Let $ p_n $ be the $n$ th prime.

I was confused about the following idea.

$$A = \sum_{n = 1}^{\infty} ( \frac{p_{2n-1}}{p_{2n}} - \frac{p_{2n}}{p_{2n +1}} ) $$

Very confused actually.

Does this even converge ? Do we need a summability method ? Does its converge depend strongly on conjectures or theorems about primes ? Does it value depend strongly on conjectures or theorems about primes ?

Is there a closed form for it ? Does its value occur in number theory ?

Does it mimic a random walk ? What is the sign of the value ?

How to use analytic number theory for this ? Or sieve theory ?

Is a zeta-regularized sum possible ?

Cesaro sum or Abel sum ?

*****Motivation*****

I assume it converges because the sums $\sum \frac{ (-1)^n}{n}$ and $\sum \frac{(-1)^n}{p_n} $ converge.

But this is apparently very different from zeta functions or prime zeta functions.

Perhaps the gaps between primes are crucial such as the twins. Or maybe arithmetic progressions. Or the largest prime gaps like Cramers conjecture or Riemann hypothesis.

Or maybe we can accelerate this infinite sum ?

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It is not an alternated series, it is obvious that its convergence depends on strong conjectures about primes.

I don't think the strongest conjecture on the prime gap even implies its convergence.

Thus the first step is to check if it converges (almost surely) when replacing $p_n$ by $n\log n$ and $p_{n+1}-p_n$ by (independent) random variables with binomial distribution of parameter $1/\log n$.