I have the below construction.
Take a standard parabola $y=x^2$ with points on the x-axis at $(-p_n,0)$ and $(p_{n+1},0)$ and corresponding points on the parabola of $(-p_n,p_n^2)$ and $(p_{n+1},p_{n+1}^2)$ respectively. The following follows regarding the point that lies on the secant intersecting the points on the parabola and the y-axis $(0,p_np_{n+1})$
$\forall p_n\in\mathbb{P},$ $p_{n}^2 \leq p_np_{n+1}$ $\implies p_{n} \leq \sqrt{p_n}\sqrt{p_{n+1}}$ $\implies p_{n}-p_{n-1} \leq \sqrt{p_n} \sqrt{p_{n+1}} - p_{n-1}$ $\implies g_{n} <\sqrt{p_n}$
Is it correct that the secant line will pass through the y-axis at the point indicated in the above construction?
Edit:
It would appear one could achieve a tighter bound by considering $(-\sqrt{g_n},0)$ and $(\frac{\sqrt{p_n}}{\sqrt{g_n}}\log({p_n}),0)$ which gives:
$g_n < \sqrt{p_n}\log{(p_n)}$
Which has some connection to the RZH by a result proved by Cramer?!
Edit2: If I change the points in the original construction to $(-\sqrt{g_n},0)$ and $(\frac{\sqrt{p_n}}{\sqrt{g_n}},0)$ the original result appears to fall out by considering the secant line point on the y-axis and the left point on the parabola.
The answer is : yes, the intersection point is $(0,p_np_{n+1})$.
Here is how one can prove it.
The equation of the line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the following condition (expressing two ways to express the slope of the line) :
$$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$$
otherwise written :
$$y-y_1=\tfrac{y_2-y_1}{x_2-x_1}(x-x_1)$$
Set now $(x_1,y_1)=(-p_n,p_n^2)$ and $(x_2,y_2)=(p_{n+1},p_{n+1}^2)$ ; you will obtain, after simplification the equation of the line under the form :
$$y-p_n^2=(p_{n+1}-p_n)(x+p_n)$$
in which you will set $x=0$ for the obtention of the value of $y$ which is indeed $p_np_{n+1}$.