Prove that $\sum_{\substack{x,y \in N \\ ax-by \ne 0 }}\frac{1}{|ax-by|xy}$ (and its generalization to higher dimensions) converges Prove that $$\sum_{\substack{x,y \in \mathbb N \\ ax-by \ne 0 }}\frac1{|ax-by|xy}$$ converges, where $a, b \in \mathbb N$
This 2-D problem can be proved by an integral test. I'm looking for some other proofs that can be easily generalized to higher dimensional cases like below,
Prove that $$\sum_{\substack{x_1,x_2, \cdots, x_d \in \mathbb N \\ \bf a \cdot \bf x \ne 0 }}\frac1{|{\bf a} \cdot {\bf x}|x_1 x_2 \cdots x_d}$$ converges, with ${\bf a} \in \mathbb Z^d$, and $\forall \; 1 \le i \le d, a_i \ne 0$.
 A: In the two dimensional case there is a simple proof. consider the set $U = \{ ax-by = 0 : x,y\in\mathbb{N}\}$. Let $A = (a,-b)$ and $v \in \mathbb{N}^2$, then for any $u\in U$, $A\cdot (v + u) = A\cdot v$. Furthermore, we have that, since $A\in\mathbb{Z}^2$, the inner product of $A$ against any vector in $\mathbb{N}^2$ is integral. So we can foliate $\mathbb{N}^2$ by lines $U_k$ such that $A\cdot u_k = k$ for $u_k \in U_k$. 
Now, 
$$ \sum_{U_k} \frac{1}{|A\cdot u| u_1 u_2} = \frac{1}{|k|} \sum_{U} \frac{1}{(x_k + u_1)(y_k + u_2)} $$
where to get to the right hand side we used the fact that on each $U_k$, as long as $a,b > 0$, we have a linear ordering of the elements by distance to the origin, and we can pick $x_k, y_k$ to be the infimum. Now, as long as $a,b \neq 0$, we have that $u_1,u_2$ are both increasing, so the inner sum is summable. Furthermore, a little bit of geometry shows you that 
$$ \sum_U \frac{1}{(x_k + u_1)(y_k + u_2)} < \frac{2}{\sqrt{x_ky_k}} \sum_U \frac{1}{u_1u_2} $$
(Edit for clarification: Using the fact that $(u_1,u_2)$ are integer multiples of a unique vector $(\hat{u}_1,\hat{u}_2)$, you have
$$ \sum_U \frac{1}{(x_k + u_1)(y_k + u_2)} \leq \frac{1}{x_ky_k}\sum_j \frac{1}{1 + j^2\frac{\hat{u}_1\hat{u}_2}{x_ky_k}} $$
[here we also need to use that $\hat{u}_1\hat{u}_2 \neq 0$] then you use the fact that 
$$ \sum_j \frac{1}{1 + \alpha j^2} \leq 1 + (\alpha)^{-1/2} \sum_k \frac{1}{1 + k^2}$$
which I leave as an exercise [if you run out of ideas, you can always prove this using integral comparison].)
The last thing to note is that $\frac{1}{x_ky_k} = O(|k|^{-1})$ using that the points $(1,k)$ and $(k,1)$ are the infimum points for the slices $U_{-bk}$ and $U_{ak}$ respectively. So 
$$ \sum_{k\neq 0} \sum_{U_k} \frac{1}{|A\cdot u| u_1u_2} \leq \sum_{k\neq 0} \frac{C}{k^{3/2}} \sum_U \frac{1}{u_1 u_2} < \infty$$

With some addition condition on what $A$ you allow, I think the claim may also be true in higher dimensions. I'll add a proof to that effect if I can think of it. 
A: Let $x_{d+1}$ denote $|a \cdot x|$, so that ${\displaystyle {1 \over |a \cdot x| x_1....x_d} = {1 \over x_1....x_{d+1}}}$. 
Then by the arithmetic-geometric inequality, for any $x_1,...,x_{d+1}$ one has
$${1 \over x_1....x_{d+1}} \leq {1 \over d+1} \sum_{i = 1}^{d+1} \prod_{j \neq i} x_j^{-{d+1 \over d}}$$
So summing one has
$$\sum_{x_1,...,x_d}{1 \over x_1....x_{d+1}} \leq {1 \over d+1} \sum_{i = 1}^{d+1} \sum_{x_1,...,x_d}
\prod_{j \neq i} x_j^{-{d+1 \over d}}$$
We examine a given term ${\displaystyle \sum_{x_1,...,x_d}\prod_{j \neq i} x_j^{-{d+1 \over d}}}$. If $i = d+1$, then one is 
adding ${\displaystyle \prod_{j = 1}^d x_j^{-{d+1 \over d}}}$ over positive integers $x_1,..x_d$ and one gets a finite result since 
${d + 1 \over d} > 1$. If $i \neq d+1$, then we sum ${\displaystyle \sum_{x_1,...,x_d}\prod_{j \neq i} x_j^{-{d+1 \over d}}}$
in the $x_i$ variable first, fixing the other variables. Only the $x_{d+1}$ 
factor depends on $x_i$, and this $x_i$ variable sum is
$$\sum_{x_i} {1 \over |a_1x_1 + ... + a_nx_n|^{d+1 \over d}} \prod_{j \neq i} x_j^{-{d+1 \over d}}$$
The sum is a portion of the sum of $|x|^{-{d+1 \over d}}$ over nonzero integers $x$, so the result is bounded by some fixed 
constant $C$. So the overall sum is at most
$$C \sum_{x_1,...,x_d}  \prod_{j \neq i} x_j^{-{d+1 \over d}}$$
In the sum above the $x_i$ variable is excluded.
Like above in the $i = d+1$ case, since the exponent ${d+1 \over d}$ is greater than one the sum is finite and we are done.
