Elements in a lattice with absolute value between two consecutive integers. Let $\Gamma$ be a lattice in $\mathbb{C}$. I am currently reading the book "A course in arithmetic" from J.P Serre, and in page 83 (for the convergence of the Eisenstein series) he claims the following:
$``$ The number of elements $\gamma \in \Gamma$ such that $|\gamma|$ is between two consecutive integers $n$ and $n+1$ is $O(n)$ $"$
Can anyone explain how this results follow? 
Thanks.
 A: The annulus $\{z\in\Bbb C\colon n\le |z|\le n+1\}$ can be covered by roughly $2\pi n$ balls of radius $1$, and any such ball can contain at most $C$ points of $\Gamma$ (for some $C$ depending on $\Gamma$).
A: There is a minimum distance $\delta$ between any two elements of a lattice in $\Bbb C$.
Therefore, for each $0 < \epsilon < 1$ and $\gamma \in \Gamma$, the closed ball $B(\epsilon\delta, \gamma)$ intersects $\Gamma$ only at $\gamma$.
Let $R(n,n+1)$ be the ring $\{z\in \Bbb C\,|\, n<|z|<n+1\}$.
For each $\gamma \in \Gamma\cap R(n,n+1)$, consider $B(\epsilon\delta, \gamma)$ and observe that the disjoin union
$$\bigcup_{\gamma \in \Gamma\cap R(n,n+1)} B(\epsilon\delta, \gamma)\subset \mathcal R = R(n-\epsilon\delta,n+1+\epsilon\delta).$$
The area of $\mathcal R$ is
$$A(\mathcal R) = \pi\left((n+1 +\epsilon\delta)^2 - (n-\epsilon\delta)^2\right) = (2n+1)(1+2\epsilon\delta)\pi.$$
Each ball $B(\epsilon\delta, \gamma)$ has area $\pi\epsilon^2\delta^2$, and hence the number of $\gamma$ that could lie in $\Gamma\cap R(n,n+1)$ is at most
$$\left\lfloor\frac{A(\mathcal R)}{\pi\epsilon^2\delta^2} \right\rfloor = \left\lfloor\frac{1+2\epsilon\delta}{\epsilon^2\delta^2}\cdot (2n+1) \right\rfloor = O(n).$$
Observe that $\delta$ is fixed and depends only on $\Gamma$ (not on $n$) and similarly $\epsilon$ is a constant.

Notice that
$$\frac{1+2\epsilon\delta}{\epsilon^2\delta^2}$$
is decreasing on $\epsilon$ for $\epsilon, \delta > 0$, so the bound is sharper for $\epsilon \to 1$.
Looking at whether $\frac{1+2\epsilon\delta}{\epsilon^2\delta^2}$ lies in $\Bbb Z$ or not and considering that the expression is decreasing on $\epsilon$, we see that we can take the bound
$$\left\lfloor\frac{1+2\delta}{\delta^2}\cdot (2n+1) \right\rfloor = O(n).$$
