TITLE: Convergence of a series related to $\mathrm{SL}_2({\mathbb N})$
QUESTION [5 upvotes]: The question The number $\pi$ and summation by $SL(2,\mathbb Z)$ considers the series
$$\sum(\lVert x\rVert+\lVert y\rVert-\lVert x+y\rVert)$$
where the series runs over vectors $x,y\in{\mathbb N}^2$ such that $\det(x,y)=1$. The norm being the standard euclidean one. Its value (equal to $2$) is obtained by a telescoping argument. However, the telescopage requires the knowledge that the auxiliary series
$$\sum\frac1{\lVert x\rVert\lVert y\rVert(\lVert x\rVert+\lVert y\rVert)}$$
converges. For this, we need to estimate the distribution of matrices  in $\operatorname{SL}_2({\mathbb N})$, in terms of the lengths of their columns.

What is known in this direction? What is approximately the number of matrices $M=(x,y)\in \operatorname{SL}_2({\mathbb N})$ such that $\lVert x\rVert+\lVert y\rVert\le R$, as $R\rightarrow+\infty$?

REPLY [3 votes]: Define $d_R(n)$ to be the number of ways to write $n=ab$ with $a$ and $b$ bounded by $R$.  Clearly, this is bounded by the standard (unrestricted by $R$) divisor function $d(n)$.  To upper-bound the desired number of matrices $M$, it suffices to bound $$\sum_{n \leq R^2} d_R(n) d_R(n+1).$$ Using only the bound $d(m) \ll m^{\varepsilon}$ shows this quantity grows at most like $R^{2+\varepsilon}$.
This can be improved slightly with more work, but maybe this is sufficient.