Convergence of $\sum_{\omega \in \Lambda\setminus \{0\}} \frac{1}{\omega^k}$ for $k > 2$ and any lattice $\Lambda\subset \Bbb C$

Question

This post is about the absolute convergence of the Eisenstein series (see these notes), i.e., for any lattice $\Lambda \subset \mathbb C$ and $k > 2$,

$$\sum_{\substack{\omega\in \Lambda\\\omega \ne 0}} \frac{1}{\omega^k}$$ converges absolutely.

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Let $\delta$ be the minimum distance between points in $\Lambda$, i.e., $$\delta := \inf\{|\lambda_1 - \lambda_2|: \lambda_1, \lambda_2 \in \Lambda\}$$ Consider the annulus $$A = A(r,\delta) := \left\{z\in \Bbb C: r \le |z| < r + \frac\delta 2 \right\}$$ $\hspace{4cm}$

Any two distinct lattice points in $A$ must be separated by an arc of length at least $δ/2$ when measured along the inner rim of $A$. It follows that $A$ contains at most $4πr/δ$ lattice points. The number of lattice points in the annulus $\{ω : n ≤ |ω| < n + 1\}$ is therefore bounded by $cn$, where $c \le (2/δ)(4πr/δ) = 8π/δ^2$. We then have $$\sum_{\substack{\omega\in \Lambda\\ \omega\ne 0}} \frac{1}{|\omega|^k} \le \sum_{n=1}^\infty\frac{cn}{n^k} = c\sum_{n=1}^\infty n^{1-k} < \infty$$ since $k > 2$.

1. Why must any two distinct lattice points in $A$ be separated by an arc of length at least $δ/2$, when measured along the inner rim of $A$? If this is true, it is clear that $A$ contains at most $\frac{2\pi r}{\delta/2} = \frac{4πr}{δ}$ lattice points.
2. Lastly, how does one see that number of lattice points in the annulus $\{ω : n ≤ |ω| < n + 1\}$ is bounded by $cn$, where $c \le (2/δ)(4πr/δ) = 8π/δ^2$? Of course, the constant $c = c(\Lambda)$ depends on the lattice.

I understand all the other details. A pictorial proof/explanation might help. Thanks a lot!