Gauss circle problem : a simple asymptotic estimation. 
Find the number of integer points lying in or inside a circle of radius $n\in \mathbb N$ centered at the origin.

The problem asks for all $(a,b)\in \mathbb Z^2$ such that $a^2+b^2\leq n^2$. Looking at such points lying strictly in the North-East quadrant, pick an $a\in \{1,\ldots,n\}$ and let $b$ run through $\{1,\ldots,\lfloor \sqrt{n^2-a^2}\rfloor\}$ : there are exactly $\displaystyle \sum_{k=1}^n \lfloor \sqrt{n^2-k^2} \rfloor$ integer points strictly in the quadrant.
Multiply that by $4$ to account for all $4$ quadrants, add the origin and points on the $x$ and $y$ axis and the total number is $$1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor$$
I'm interested in asymptotics of this sum. A trivial estimate is the following:$$\begin{align}\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor &= \sum_{k=0}^n \sqrt{n^2-k^2} + O(n)\\ &= n^2\cdot \left(\frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}}\right) + O(n) \\ &=n^2 \left( \int_0^1 \sqrt{1-t^2}dt + o(1)\right) + O(n) \\ &= \frac \pi 4n^2 + o(n^2)\end{align} $$
Hence $$1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor = \pi n^2 + o(n^2)$$

Is there a way to refine the estimate above and get $\pi n^2 + O(n)$ instead (or something better) ? 

 A: Pick a square with unit side length centered at every lattice point inside the region $x^2+y^2=n^2$, i.e. the circle with radius $n$ centered at the origin. Those squares entirely cover the circle with radius $n-\frac{1}{\sqrt{2}}$, but any square lies inside a circle with radius $n+\frac{1}{\sqrt{2}}$, hence the number of lattice points is between $\pi\left( n-\frac{1}{\sqrt{2}}\right)^2$ and $\pi\left(n+\frac{1}{\sqrt{2}}\right)^2$, so it is $\pi n^2+O(n)$.
For info about the Voronoi bound, please see this survey.
A: The estimate $\displaystyle \frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}} = \int_0^1 \sqrt{1-t^2}dt + o(1)$ can easily be refined .
Indeed, $$\begin{align}0\leq \frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}} - \int_0^1 \sqrt{1-t^2}dt &= \sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n} \left(\sqrt{1-\frac{k^2}{n^2}} - \sqrt{1-t^2}\right)dt\\ &\leq\sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n} \left(\sqrt{1-\frac{k^2}{n^2}} - \sqrt{1-\frac{(k+1)^2}{n^2}}\right)dt  \\ &= \frac 1n \end{align}$$
Hence $\displaystyle \frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}} = \int_0^1 \sqrt{1-t^2}dt + O(\frac 1n)$ and $$1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor = \pi n^2 + O(n)$$
