It is known that the Riemann zeta function
$$\displaystyle \zeta(s) = \sum_{n = 1} \frac{1}{n^s}$$
converges for all $\mathrm{Re}(s) > 1$, and admits an analytic continuation to the rest of the complex plane apart from its poles. My question concerns a generalisation of this: suppose that we replace $n$ with a vector $m \in \mathbb{Z}^d$ and consider the sum
$$\displaystyle \sum_{\substack{\ \ m \in \mathbb{Z}^d} \\ {\ \ \ m \neq 0}} \frac{1}{|m|^s}.$$
Does this sum also converge absolutely for $\mathrm{Re}(s) > 1$? It clearly converges if $d = 1$. If $d = 2$, then we have:
$$\displaystyle \sum_{\substack{\ \ m_1 \in \mathbb{Z}} \\ {\ \ \ m_1 \neq 0}} \sum_{\substack{\ \ m_2 \in \mathbb{Z}} \\ {\ \ \ m_2 \neq 0}} \frac{1}{|m_1 + m_2|^s}, \ \ m_1 + m_2 \neq 0.$$
It doesn't seem clear to me why this should converge. The closest I've been able to find is the so-called Hurwitz zeta function, which looks somewhat similar to my sum except that I'm summing over the parameter $q$ as well. I've tried comparing the double sum to a double integral of the same thing over $\mathbb{R}^2$, but Mathematica tells me that the resulting integral does not converge for $s = 3/2$, which is not a good sign.
Using Mathematica on the sum directly -- summing over $\mathbb{N}$ rather than $\mathbb{Z}$ -- yields $\zeta(s - 1) + \zeta(s)$ for the case $d = 2$.
use the change of variable $r = x+y$ for showing that $$\iint_{[1,\infty)\times[1,\infty)} f(x+y)dxdy = \int_2^\infty (r-2)\, f(r)dr$$ Then we can do the usual comparison with a Riemann integral $$\sum_{n,m \in \mathbb{N}^*\times \mathbb{N}^*} |n+m|^{-s}= \iint_{[1,\infty)\times[1,\infty)} (\lfloor x\rfloor+\lfloor y\rfloor)^{-s} dxdy$$ $$\iint_{[1,\infty)\times[1,\infty)} (x+y)^{-s}dxdy +\iint_{[1,\infty)\times[1,\infty)} (\lfloor x\rfloor+\lfloor y\rfloor)^{-s}-(x+y)^{-s}) dxdy$$ $$ = \int_2^\infty (r-2)r^{-s}dr + \iint_{[1,\infty)\times[1,\infty)} \mathcal{O}((x+y)^{-s-1})dxdy =\int_2^\infty (r-2)r^{-s}dr + \int_1^\infty \mathcal{O}(r^{-s})dr$$ where I used that $\displaystyle(\lfloor x\rfloor+\lfloor y\rfloor)^{-s} - (x+y)^{-s} = s \int_{\lfloor x\rfloor+\lfloor y\rfloor}^{x+y} t^{-s-1}dt = \mathcal{O}((x+y)^{-s-1})$
The same argument shows that $\displaystyle\sum_{(n_1, \ldots, n_d) \in (\mathbb{N}^*)^d} (n_1+\ldots +n_d)^{-s}$ converges for $Re(s) > d$.