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 $(2)$?

  • 1
    $\begingroup$ 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. $\endgroup$
    – K.defaoite
    Commented Sep 21, 2020 at 10:17
  • 1
    $\begingroup$ 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! $\endgroup$ Commented Sep 21, 2020 at 13:14


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