Q:

When does the sum $\sum_{k\in\mathbb{Z}^n}\langle k\rangle^{-s}$ converge?

I'm stuck with the following problem: determine $s\in\mathbb{R}$ so that the sum
$$
\sum_{k\in\mathbb{Z}^n}\langle k\rangle^{-s}, \quad \langle k\rangle=(1+|k|^2)^{1/2}
$$
converges. My first attempt was to convert this series into an integral
$$
\sum_{k\in\mathbb{Z}^n}(1+|k|^2)^{-s/2}\rightarrow\int_{\mathbb{R}^n}\frac{\text{d}x}{(1+|x|^2)^{s/2}}
$$
and when using spherical coordinates we get
$$
\int_{\mathbb{R}^n}\frac{\text{d}x}{(1+|x|^2)^{s/2}}=C\int_0^{\infty}\frac{r^{n-1}}{(1+r^2)^{s/2}}\text{d}r
$$
where $C>0$ is the measure of the unit ball in $\mathbb{R}^n$. The integrand in the last integral behaves like $r^{n-1-s}$ when $r\rightarrow\infty$. Therefore the integral is finite iff $s>n$. 
My question is: can I justify somehow the step where I change the series into an integral? Is there a multidimensional version of the integral test and what does it assume from the function? In 1D-case we need the function to be continuous and decreasing.

A:

Strictly to the finally stated question: Yes. 
To have an easy comparison (and know the control direction in inequalities), let us consider $s>n$ first, show the convergence with the argument in the OP.
One can consider and analyze only the sum over $k\in\Bbb Z_{\ge 0}^n$ (since each $k=(k_1,k_2,\dots)$ contributes to the sum via $k_1^2,k_2^2,\dots$), then inductively arguing even over $k\in\Bbb Z_{> 0}^n$ by eliminating one by one the sums over $k_1=0$, $k_2=0$, ... since $s>n>n-1$.
Now associate for each $k>0$ (inequality on components) the cube $C_-(k)$ between $k$ and $k-\underline 1$. Here $\underline 1=(1,1,\dots)$, and $C_-(k)=\{\ x\ :\ x_1\in[k_1-1,k_1]\ ,\ x_2\in[k_2-1,k_2]\ ,\ \dots\ \}$.
Then we can estimate:
$$
\frac 1{(1+|k|^2)^{s/2}}
=
\int_{C_-(k)}\frac 1{(1+|k|^2)^{s/2}}\; dx
\le
\int_{C_-(k)}\frac 1{(1+|x|^2)^{s/2}}\; dx\ .
$$
On the other side, e.g. for $s=n$ we have to use the "other cube",
$C_+(k)$ between $k$ and $k+\underline 1$, $C_+(k)=\{\ x\ :\ x_1\in[k_1,k_1+1]\ ,\ x_2\in[k_2,k_2+1]\ ,\ \dots\ \}$.

