Some time ago I found this sum of prime numbers converge(*)

$$\sum_{k=1}^\infty \frac{p(k+1)-2p(k+2)+p(k+3)}{p(k)-p(k+1)+p(k+2)}\ \approx \frac{5}{7}\zeta(3)^{-2} $$ where:

$\zeta(s)$ is the Riemann zeta function

$p(n)$ is the $n^{th}$ prime number

Working a bit on I found that: $$\sum_{k=1}^n \frac{p(k+1)-p(k+2)+p(k+3)-p(k+4)+p(k+5)}{p(k)-p(k+1)+p(k+2)-p(k+3)+p(k+5)}\ \approx n $$

this means that $$F(k)=\frac{p(k+1)-p(k+2)+p(k+3)-p(k+4)+p(k+5)}{p(k)-p(k+1)+p(k+2)-p(k+3)+p(k+5)}\approx 1$$

$F(k):$

$F(k)=1 \pm \varepsilon_{n}$

$p(k+1)-p(k+2)+p(k+3)-p(k+4)+p(k+5)=p(k) \pm \varepsilon_{n}p(k)-p(k+1)\pm \varepsilon_{n}p(k+1)+p(k+2)\pm \varepsilon_{n}p(k+2)-p(k+3)\pm \varepsilon_{n}p(k+3)+p(k+5)\pm \varepsilon_{n}p(k+5).$

$E_k=\pm\varepsilon_n(p(k)+p(k+1)+p(k+2)+p(k+3)+p(k+5))$

$ \ $

### $$p(k+4)=2p(k+3)-2p(k+2)+2p(k+1)-p(k)+E_{k}$$

$E_{k}:$

In this way you can look for the next prime number taking into account the error $E_{k}$.

How can I calculate the range of $E_{k}$?

I do not know if this way has already been used by someone, I had the pleasure of sharing it because of my lack of mathematical knowledge I would not know how to go on