Does this sum of prime numbers converge? $\newcommand{\P}{\operatorname{P}}$I'm wondering if this sum of prime numbers converges and how can I estimate the value of convergence.
$$\sum_{k=1}^\infty  \frac{\P[k+1]-2\P[k+2]+\P[k+3]}{\P[k]-\P[k+1]+\P[k+2]}$$
$ \ $
\begin{align}
\sum_{k=1}^{10}  \frac{\P[k+1]\cdots}{\P[k]\cdots} & = 0.4380952380952381` \\
& \,\,\, \vdots \\
\sum_{k=1}^{10^5}  \frac{\P[k+1]\cdots}{\P[k]\cdots} & =0.49433323447491884` \\[10pt]
\sum_{k=1}^{10^6}  \frac{\P[k+1]\cdots}{\P[k]\cdots} & = 0.49433634247938607`\ \approx \frac{5}{7}\zeta(3)^{-2}
\end{align}
$\ $ 
$ \zeta(s) \ $ is the Reimann zeta function
P$[n] \ $ is the $n^\text{th}$ prime number
$\ $ 

This is enough to assert that the series converges?
How can I estimate the value of convergence and if it is rational or irrational?
 A: Letting $b_k = p_k - p_{k+1}+p_{k+2}$
and $a_k = g_{k+2} - g_{k+1}$ where $g_k = p_{k+1}-p_k$, 
summing by parts
$$\sum_{k=1}^n \frac{a_k}{b_k} = \frac{g_2-g_{n+2}}{b_n}+\sum_{k=1}^{n-1} (g_2-g_{k+2})(\frac{1}{b_k}- \frac{1}{b_{k+1}})$$
Since $b_k \sim k \log k$ $$\frac{1}{b_k}- \frac{1}{b_{k+1}}= \frac{b_k-b_{k+1}}{b_kb_{k+1}}\sim  \frac{g_k+g_{k+1}+g_{k+2}}{k^2 \log^2 k}$$
Also we know $g_k = \mathcal{O}(k^\theta)$ for some $\theta < 0.6$ but this is not sufficient to conclude.

We look instead at
$$\sum_{k=2}^\infty \frac{g_k g_{k+c}}{k^2 \log^2 k} \overset{?}< \infty \tag{1}$$
By summation by parts
$\sum_{k=2}^K  \frac{g_{k+c}}{k \log k}  \sim \log K$ so I'd say yes $\sum_{k=2}^\infty \frac{g_k g_{k+c}}{k^2 \log^2 k}$ converges, but I can't prove it.
Summing by parts $\sum_{k=K}^\infty \frac{g_k }{k^2 \log^2 k} \sim \frac{1}{k \log k}$ lets me think $(1)$ converges.
A: As I commented on @user1952009's answer, the series converges under assumption of Cramer's conjecture. However, we can prove the convergence of the series unconditionally. We have in fact a result on partial sums of squares of prime gaps which was proven by R. Heath-Brown. Here's the link to the paper: 

Theorem [Heath-Brown] 
Let $p_n$ be the $n$-th prime, and let $g_n  = p_{n+1}-p_n$. Then we have
  $$
\sum_{n\leq x} g_n^2 \ll x^{\frac{23}{18}+\epsilon}.
$$

After @user1952009's answer, it suffices to consider the convergence of (1): which is 
$$
\sum_{k=2}^{\infty} \frac{g_k g_{k+c}}{k^2 \log^2 k} < \infty.
$$
Note that $2g_kg_{k+c}\leq g_k^2 + g_{k+c}^2$, so it suffices to consider the convergence of 
$$
\sum_{k=2}^{\infty} \frac{g_k^2}{k^2\log^2 k} 
$$
since the same idea will apply to $\sum_{k=2}^{\infty} \frac{g_{k+c}^2}{k^2\log^2 k}$.
Let $A(x)=\sum_{k\leq x} g_k^2$, $f(x) = \frac1{x^2 \log^2 x}$. Heath-Brown's result states $A(x) \ll x^{23/18+\epsilon}$. Then by partial summation, we have
$$
\begin{align}
\sum_{2\leq k\leq x} \frac{g_k^2}{k^2\log^2 k}&=\int_{2-}^x f(t) dA(t) \\
&=f(t)A(t) \bigg\vert_{2-}^x -\int_{2-}^x A(t) f'(t)dt\\
&=f(2-)A(2-) + O\left( x^{-\frac{13}{18}+\epsilon} \right)+\int_{2-}^x  \frac{2A(t) (\log t + 1) }{t^3\log^3 t}dt
\end{align}
$$
Now, we have the convergence since $23/18 - 3 = -31/18<-1$.
Another problem that uses Heath-Brown's result is here.
