Proving convergence of sum over $\mathbb{Z}^n$

Question

In my research, I am trying to use the following construction by Benson Farb and John Franks, which proves that for all $n$, the group of $n\times n$ matrices with 1's on the diagonal, 0's above the diagonal and integer entries below embeds as a subgroup of $C^1(S^1)$.

http://www.math.uchicago.edu/~farb/papers/nilpotent.pdf

The construction hinges on defining interval lengths with the following sums:

Let $K>0$ and $B_K : \mathbb{Z}^n \to \mathbb{R}$ be defined by \begin{align*} B_K(q_1,q_2,\ldots,q_n) &= K + \sum_{j=1}^{n} q^{2n-2j+2}_j\\ &=q^{2n}_1+q^{2n-2}_2+\cdots+q^4_{n-1}+q^2_n +K \end{align*} and let $S_K$ be defined by $$ S_K = \sum_{(q_1,q_2,\ldots,q_n)\in\mathbb{Z}^n} \frac{1}{B_K(q_1,q_2,\ldots,q_n)}.$$ The authors off-handedly say the sum defining $S_K$ converges by the integral and comparison tests. When I first saw the sum, I was like "Pfff of course, I'll just do an inductive comparison to the harmonic series. Easy squeezy lemons."

Two days later and I have no idea how to prove convergence.

I've tried induction on $n$. Clearly converges for $n=1$, assume true for $n-1$ and write \begin{align*} S_K =\sum_{q_1\in\mathbb{Z}}\ \sum_{(q_2, q_3,\ldots,q_n)\in\mathbb{Z}^{n-1}} \frac{1}{q^{2n}_1+q^{2n-2}_2+\cdots+q^4_{n-1}+q^2_n +K}\end{align*} but I cannot figure out a way to extract a $q_1$ term to prepare for the second summation.

I've also tried to sum one $q_i$ at a time, performing the integral test at each step, i.e.

$$S_K=\sum_{(q_1, q_2,\ldots,q_{n-1})\in\mathbb{Z}^{n-1}}\ \sum_{q_n\in\mathbb{Z}} \frac{1}{A + q^2_n}$$

where $A$ is all of the other terms. Then, by the integral test $$S_K \leq \sum_{(q_1, q_2,\ldots,q_{n-1})}\frac{1+\frac{\pi}{2}\sqrt{A}}{A}$$ but this square root stirs up trouble for me because I'm forced to halve the exponents of all the terms in $A$ for my next approximation. This works for $n=2$ and $n=3$, but I can't push it further.

I'm close to foaming at the mouth, so if y'all have any input it would be greatly appreciated.

Answer 1

The sum diverges already for $n=4$. To see this, let $L\geq K$ be a dyadic parameter, and consider the contribution of $q_1\asymp L^{1/8}$, $q_2\asymp L^{1/6}$, $q_3\asymp L^{1/4}$, $q_4\asymp L^{1/2}$. If the implied constants are sufficiently close to each other, these ranges are pairwise disjoint. However the contribution of such a range is $\asymp L^{1/8+1/6+1/4+1/2}/L>1$. Summing up over the various $L$'s, the claim follows.

Answer 2

The series indeed diverges for $n\ge4$. Indeed, without loss of generality $K\ge1$. For real $A\ge1$ and $p>1$ we have
\begin{equation} \int_1^\infty\frac{dq}{A+q^p}\ge\int_1^\infty\frac{dq}{(A^{1/p}+q)^p} \gg\frac1{A^{1-1/p}}. \end{equation} So, for $n\ge4$ \begin{align*} S_K&\ge\int_{[1,\infty)^n} \frac{dq_1\cdots dq_n}{K+q^{2n}_1+\cdots+q^{8}_{n-3}+q^{6}_{n-2}+q^4_{n-1}+q^2_n} \\ &\gg\,\int_{[1,\infty)^{n-1}} \frac{dq_1\cdots dq_{n-1}}{K+q^{n}_1+\cdots+q^{4}_{n-3}+q^{3}_{n-2}+q^2_{n-1}} \\ &\gg\,\int_{[1,\infty)^{n-2}} \frac{dq_1\cdots dq_{n-2}}{K+q^{n/2}_1+\cdots+q^{2}_{n-3}+q^{3/2}_{n-2}} \\ &\gg\,\int_{[1,\infty)^{n-3}} \frac{dq_1\cdots dq_{n-3}}{K+q^{n/6}_1+\cdots+q^{2/3}_{n-3}} =\infty. \end{align*}

Answer 3

The authors just posted a correction to the paper on arXiv, where they say:

"In an earlier version of this paper we used $2n − 2j + 2$ as the exponent in the definition of BK instead of $4n − 2j + 2.$ As a result (as was pointed out to us) the series SK might not converge. However, the only properties we use of BK are that SK converges and that the ratio of the value of BK at certain points limits to 1."