$$\sum_{k=1}^{n}\left\lfloor\sqrt{ r^2-k^2 }\right\rfloor$$
where $r$ is a constant (not necessarily an integer). Note that $r\ge n$ and both $r, n$ are positive.
I apologize if I'm not able to contribute a lot to this problem as the floor function is really weird to me. I have never seen this inside a summation and don't know what to make of this problem. If anything, can one at least place bounds on this summation?
I know that the floor function is defined to satisfy the following property:
If $x:=\left\lfloor n \right\rfloor$, then
$$x-1<n\le x$$
How should I proceed? Are there any properties that I should know about when dealing with floor functions in summations?
UPDATE: Thank you for the answer. Are there bounds that I can place around the summation that get more accurate as $n$ gets larger?
Consider the $xy$-plane, which contains positive lattice points $(j,k)$ with $j,k\in\Bbb N$. For any fixed $k$, the expression $\lfloor\sqrt{r^2-k^2}\rfloor$ counts the number of positive integers $j\le\sqrt{r^2-k^2}\rfloor$, which is the same as $j^2+k^2\le r^2$. In other words, $\lfloor\sqrt{r^2-k^2}\rfloor$ is the number of positive lattice points at height $k$ inside the circle of radius $r$ centered at the origin. The entire sum is therefore the number of positive lattice points of height at most $n$ inside that circle.
There isn't a nice closed formula for that quantity, even when $n\ge r$ so that the height restriction is unnecessary. But it is a standard result that the number of lattice points of a "nice" region is approximately its area. So the number of lattice points will be approximately the area of the region $\{(x,y)\in\Bbb R^2\colon x\ge0,\, 0\le y\le n,\, x^2+y^2\le r^2\}$, which is (fun calculus problem) $$ \begin{cases} \displaystyle \frac{1}{2} \left(n \sqrt{r^2-n^2}+r^2 \arctan\left(\frac{n}{\sqrt{r^2-n^2}}\right)\right), &\text{if } n\le r, \\ \dfrac{\pi r^2}4, &\text{if } n\ge r. \end{cases} $$