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Let be $ \Lambda\subseteq \mathbb C$ a lattice, I don't understand why the series $$\sum_{\lambda\in\Lambda\setminus\{0\}} \frac{1}{|\lambda|^s}$$ converges for $s>2$. Can someone help me?

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Let $D$ be a findamental region of $\Lambda$ centered at the origin. Now we can find positive constants $C$ and $c$, depending only on $s$ such that for each $\lambda\in\Lambda-\{0\}$, $c|x|^s\leq|\lambda|^s\leq C|x|^s$ on each $\lambda+D$. This enables us to deduce that the summation in question is comparable to $\int_{\mathbb{R}^2-D}|x|^{-s}\;dx$. Examining the convergence region for this integral is easy, hence the conclusion follows. – Sangchul Lee Aug 3 '12 at 13:00
up vote 5 down vote accepted

Start by looking initially, the the first 8 terms (with $\lambda$ nearest to $0$), in terms of the smallest distance of these 8 values of $\lambda$ from $0$ (call it $r$). You should get $$\frac{1}{|\lambda^s|} \leq \frac{1}{r^s}.$$

Then for the next 16 terms (slightly further away from $0$), you should be able to get: $$\frac{1}{|\lambda^s|} \leq \frac{1}{(2r)^s}$$ and so on for values of $\lambda$ further out from $0$.

If you then add the first $8(1+2+\dots+n)$ values of $\lambda$, you find that the partial sum is bounded above by $$\frac{8}{r^s}\sum_{k=1}^n \frac{1}{k^{s-1}},$$ which should give the convergence for $s>2$. You could also look at lower bounds in the same way if you wanted to show that the series converges if and only if $s>2$.

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