# Number of 'integer unit squares' on and inside a circle

Let $$b(R)$$ and $$s(R)$$ be the number of integer unit cubes in $$\mathbb{R}^m$$ that intersect the ball and sphere of radius $$R$$, centered at the origin. If $$m=2$$, calculate the limits $$\lim_{R\rightarrow \infty}\frac{s(R)}{b(R)} \quad\text{ and }\quad\lim_{R\rightarrow \infty}\frac{s(R)^2}{b(R)}.$$

I believe the first limit is 0, but I'm stuck on the second one.

My approach:

Calculate $$s(R)$$ and $$b(R)$$ in the first quadrant to make things easier.

The number of unit squares of $$b(R)$$ in the first quadrant lying between the lines $$x=n$$ and $$x=n+1$$ (for $$0\le n\le \lfloor R\rfloor$$) is $$\lceil\sqrt{R^2-n^2}\rceil$$ (apply Pythagoras Theorem), so $$b(R)/4= \sum_{n=0}^{\lfloor R\rfloor} \lceil \sqrt{R^2-n^2}\rceil.\tag{1}$$

Working similarly as above, we find the number of squares of $$s(R)$$ in the first quadrant lying between the lines $$x=n$$ and $$x=n+1$$ to be $$\lceil\sqrt{R^2-n^2}\rceil - \lfloor\sqrt{R^2-(n+1)^2}\rfloor$$. Summing up for all columns, we find a nearly telescopic sum which can be at most $$2\lfloor R\rfloor +1$$. So $$s(R)/4 \le 2\lfloor R\rfloor +1.\tag{2}$$ We easily get $$\frac{s(R)}{b(R)}\le \frac{2\lfloor R\rfloor +1}{\sum_{n=0}^{\lfloor R\rfloor} \lceil \sqrt{R^2-n^2}\rceil}\le \frac{2\lfloor R\rfloor+1}{\sum_{n=0}^{\lfloor R\rfloor}\lfloor R\rfloor-n}.\tag{3}$$ The denominator of the last expression is a quadratic in $$\lfloor R\rfloor$$, so $$\lim_{R\rightarrow \infty} \frac{s(R)}{b(R)}=0.$$

How can I find the second limit using $$(1)$$ and $$(2)$$?

• I'm not sure how helpful this is, but a very similar topic in dimension 2 is discussed well in this Mathologer YouTube video. Obviously the video is designed for a somewhat wider audience, but I think it still presents some good ideas quite well. Commented Sep 21, 2020 at 10:17
• Thanks, I liked the video a lot :D I'm a bit in the dark as to how I can proceed with that idea, but nevertheless I'll give it a go! Commented Sep 21, 2020 at 13:14