That's a Parabola!

Let $\theta = \angle OFB$ where $B(x, \frac{x^{2}}{4})$ lies on the given parabola and let $L$ be the length of $FB$ . We want to find an expression for $L$ in terms of $\theta$ ,

Then $x = L\sin(\theta)$ and, since $L$ is the distance between $F(0,1)$ and $B$ , we have that

$L^{2} = x^{2} + (1 - \frac{x^{2}}{4})^{2} = \frac{x^{4}}{16} + \frac{x^{2}}{2} + 1 = (\frac{1}{16})(x^{4} + 8x^{2} + 16) = (\frac{1}{16})(x^{2} + 4)^{2}$ ,

and so $L = (\frac{1}{4})(x^{2} + 4)$ . Now substitute in $x = L\sin(\theta)$ to find that

$4L = L^{2}\sin^{2}(\theta) + 4 \Longrightarrow L^{2}\sin^{2}(\theta) - 4L + 4 = 0$ .

Now use the quadratic equation to find that

$L = \dfrac{4 \pm \sqrt{16 - 16\sin^{2}(\theta)}}{2\sin^{2}(\theta)} = \dfrac{2 \pm 2\cos(\theta)}{1 - \cos^{2}(\theta)} = \dfrac{2}{1 + \cos(\theta)} = \sec^{2}(\frac{\theta}{2})$ ,

where I took the negative root so that $L = 1$ for $\theta = 0$ . So the limit we are looking at is of the form

$S = \lim_{n\rightarrow \infty} \dfrac{1}{n} \displaystyle\sum_{k=1}^{n} \sec^{2}(\frac{k\pi}{4n})$ .

Now I haven't seen the sum of $sec^{2}(\theta)$ before so I had to resort to using WolframAlpha to determine this limit numerically to obtain a value of $S = \boxed{1.2733}$ to $4$ decimal places.