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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 $\forall \; 1 \le i \le d, a_i \ne 0$.

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    a = 100, b=10, x=1, y=10. Please put some more effort into forming the question. Also, what have you tried? Where are you stuck? Is this homework by any chance?2010-12-11
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    thanks for the remind. I re-expressed the problem.2010-12-13
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    Your claim is false in general. Consider $A = (1,0,\ldots,0)$, the summand becomes $(x_1^2x_2\cdots x_d)^{-1}$ is not summable on $\mathbb{N}^d$. Even in two dimensions you need to use rather crucially the fact that $a,b \neq 0$.2010-12-13
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    You are right. I added a constraint to $\bf A$ so that no element of it can be zero. for 2-D case $a,b \in N \, $ is ok.2010-12-13

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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.

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    Small correction, in the middle paragraph, "If $i = d$" and later "if $i \neq d$" should have $d+1$ on the RHS. This is a much more satisfactory proof than mine. (It didn't occurr to me to apply the AM-GM to also the $a\cdot x$ term. Though on hindsight, it should.)2010-12-15
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    And of course, your approach implies also the integral version of the result: "Let $\{v_1, \ldots, v_{d+1}\}$ be a set of vectors in $\mathbb{R}^d$ such that any subset of $d$ elements is linearly independent. Then the integral $$ \int_{\mathbb{R}^d} \prod_{i = 1}^{d+1} (1 + |v_i\cdot x|)^{-1} dx $$ is finite".2010-12-15
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    Thanks, corrected that. For the integral versions, you do have to cut the integrals off away from the origin.2010-12-15
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    Actually come to think of it you'd have to cut off each factor near where it blows up to avoid divergence.2010-12-15
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    Thanks, Zarrax. A very good answer. But I guess the exponential $- \frac{d+1}{d}$ should be $- \frac{d+2}{d+1}$.2010-12-16
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    @Zarrax: that is why the product contains the factor $(1+\cdots)$. :) @user4649: I think Zarrax's exponential works. Consider replacing $x \to \lambda x$ for positive $\lambda$. Zarrax's expression scales homogeneously on the left and right hand sides.2010-12-17
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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.

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    Can you explain a bit more on how "a little bit of geometry shows you that, $$ \sum_{U} \frac{1}{(x_k+u_1)(y_k+u_2)} < \frac{1}{x_ky_k} \sum_U \frac{1}{u_1u_2} $$ " I found it not very obvious.2010-12-14
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    You are right. It is not obvious because as I wrote it it was wrong. Sorry about being somewhat sloppy. There should be a square-root in the denominator, and a factor of 2 in the numerator. You need to use the fact that $u_1u_2$ grows quadratically. I added a few lines to clarify that. I hope it is better now.2010-12-14
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    Thanks, Willie. This is also a good answer2010-12-16