Sum of a series in $\mathbb{Z}^d$ Regarding the question, where it was answered that $$\left|\int_{[-1,1]^d}e^{ik\cdot x}dx\right|\lesssim\frac{1}{\prod_{j=1}^d(1+|k_j|)},$$ the paper in question also claims that the sum
$$\sum_{k\in\mathbb{Z}^d}\frac{1}{\prod_{j=1}^d(1+|k_j|)^2}$$ is finite. I am not sure how to prove this. I tried to partition $\mathbb{Z}^d$ into sets indexed by a single parameter rather than a multi-parameter so that the usual series convergence tests could be applied, but there does not seem to be a natural way to partition $\mathbb{Z}^d$, other than radially, for example.
Any hints or ideas would be greatly welcomed. 
 A: Fix $k_1, \ldots, k_{d-1}$. Then 
$$\sum\limits_{k_d \in \mathbb{Z}}\dfrac{1}{\prod\limits_{j=1}^d(1 + |k_j|)^2} = \dfrac{1}{\prod\limits_{j=1}^{d-1}(1 + |k_j|)^2}\sum\limits_{k_d\in\mathbb{Z}}\frac{1}{(1 + |k_d|)^2}.$$
That final sum converges, since it's just $$2\sum\limits_{k>0}\frac{1}{k^2} - 1.$$
Now, we can do the same for all of the other coordinates and pull out the factors, giving
$$\sum\limits_{k\in\mathbb{Z}^d}\dfrac{1}{\prod\limits_{j=1}^d(1+|k_j|)^2} = \prod\limits_{i = 1}^d\left(\sum\limits_{k_i\in\mathbb{Z}}\frac{1}{(1 + |k_d|)^2}\right).$$
Since all of those sums converge, so does their product. But since we've gone this far, we might as well finish it off and calculate exactly what it converges to: it's
\begin{align*}\sum\limits_{k\in\mathbb{Z}^d}\dfrac{1}{\prod\limits_{j=1}^d(1+|k_j|)^2} &= \prod\limits_{i = 1}^d\left(\sum\limits_{k_i\in\mathbb{Z}}\frac{1}{(1 + |k_d|)^2}\right)\\&=\prod\limits_{i=1}^d\left(2\sum\limits_{k>0}\frac{1}{k^2} - 1\right)\\&=\prod\limits_{i=1}^d\left(\frac{\pi^2}{3} - 1\right)\\&=\sum\limits_{i=1}^d\left(\frac{\pi^2}{3}\right)^i(-1)^{d-i}\left(\array{d\\i}\right).\end{align*}
