<s>[INST] Convergence of $1/(x^2+y^2)^a$. I'm curious about the convergence of the series:
\begin{align}
 \sum_{x,y=1}^\infty \frac{1}{(x^2+y^2)^\alpha}\ ,\ \alpha \in \mathbb{N}
\end{align}
I'm wondering for what values of $\alpha$ this converges, and if so is there an analytic way to find the limit? I'm not really sure where to start, it's been a long time since I studied sequences and series. 
 [/INST] Let $N,M \in \Bbb N$. On the square $[N,N+1]\times [M,M+1]$,
$$\frac{1}{[(N+1)^2 + (M + 1)^2]^\alpha} \le \frac{1}{(x^2 + y^2)^\alpha} \le \frac{1}{(N^2 + M^2)^\alpha}.$$
Integrating the inequality over the square, we find
$$\frac{1}{[(N+1)^2 + (M+1)^2]^\alpha} \le \int_N^{N+1}\int_M^{M+1} \frac{dx\, dy}{(x^2 + y^2)^{\alpha}} \le \frac{1}{(N^2 + M^2)^\alpha}.$$
Summing from over all $N,M \ge 1$, we have
$$\sum_{N,M\ge 2} \frac{1}{(N^2 + M^2)^\alpha} \le \int_1^\infty \int_1^\infty \frac{dx\, dy}{(x^2 + y^2)^\alpha} \le \sum_{N,M\ge 1} \frac{1}{(N^2 + M^2)^\alpha}$$
Hence, the series $\sum_{x,y\ge 1} (x^2 + y^2)^{-\alpha}$ converges if and only if $\int_1^\infty \int_1^\infty (x^2 + y^2)^{-\alpha}dx\, dy$ converges. Using the transformation $x = u\tan \theta$, $y = u$, we obtain
$$\int_1^\infty \int_1^\infty \frac{dx\, dy}{(x^2 + y^2)^\alpha} = C_\alpha\int_1^\infty u^{1 - 2\alpha}\, du $$
where $C_\alpha = \int_{\pi/4}^{\pi/2} \cos^{2\alpha - 2}\theta\, d\theta$. Since $C_\alpha$ is finite (since $\alpha \ge 1$) and $\int_1^\infty u^{1 - 2\alpha}\, du$ converges if and only if $\alpha > 1$, it follows that $\sum_{x,y \ge 1} (x^2 + y^2)^{-\alpha}$ converges if and only if $\alpha > 1$.
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