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.