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.



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


