I am interested in the asymptotic behavior of $$S_{\beta}(\mu) := \sum_{0<|n|\leq \mu} \big(|n|^{-2\beta} - \mu^{-2\beta}\big)^{\frac12} \sim ~ ?~,~ \mu \to \infty $$ where the sum runs over the $n = (n_1,n_2) \in \mathbb{N}^2$. The distance is the usual Euclidean distance in $\mathbb{R^2}$: $$|(n_1,n_2)|=(n_1^2 + n_2^2)^{\frac12}$$ When $\beta > 2$, the serie $\sum_{0 < |n|} |n|^{-\beta}$ converges, thus, by dominated convergence the asymptotic is a constant.
I would like to know what happens when $1 < \beta \leq 2$.
I tried to use a kind of serie/integral comparison test, but I struggle to properly justify it, and I don't know if it is true. Moreover the integral I obtain seems, at least to me, quite complicated : $\int \int_{|x| \leq \mu} \big(|x|^{-2\beta} - \mu^{-2\beta}\big)^{\frac12} dx$.
I am really familiar with usual serie $\sum_{n \geq 0} u_n$ but not with "double" series. Even the asymptotic of $\sum_{0<|n|\leq \mu} |n|^{-\beta}$ when $1 < \beta \leq 2$ would help me. Especially the method to obtain it.
Thanks,
EDIT :
Thanks to Gary's comment, I gathered the terms by $n_1^2+n_2^2 = k$ where $0 < k \leq \mu^2$. So, assuming that $\mu$ is an integer :
$$S_{\beta}(\mu) = \sum_{k = 1}^{\mu^2} \sum_{n_1^2+n_2^2 = k} \big(k^{-\beta} - \mu^{-2\beta}\big)^{1/2} = \sum_{k = 1}^{\mu^2} r_2(k) \big(k^{-\beta} - \mu^{-2\beta}\big)^{1/2}$$
where $r_2(n)$ is the number of decompositon of $n$ as the sum of two square numbers (mathworld.wolfram.com/SumofSquaresFunction.html).
I think that it is easier to get properties of "means" of $r_2$ so, by Abel transform :
$$S_{\beta}(\mu) = \sum_{k = 1}^{\mu^2-1} R_k u_k$$
with $R_k = \sum_{j = 1}^k r_2(j)$ and $u_k = \big(k^{-\beta} - \mu^{-2\beta}\big)^{1/2}- \big((k+1)^{-\beta} - \mu^{-2\beta}\big)^{1/2}$
I read that $\sum_{j = 1}^k r_2(j) = \pi k + O(\sqrt{k})$. So, I think I the asymptotic behavior is $\mu^{2-\beta}$.