Q:

Convergence of double series $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{\sin(\sin(nm))}{n^2+m^2}$

I was playing around with some integrals and series convergence and computations and after some ugly transformations the following double series occurred. Title says it all, is the following series convergent or divergent? If its convergent can we get a good estimate?
$$\displaystyle{ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{\sin(\sin(nm))}{n^2+m^2}}.$$
An elementary solution is preferred.
Thanks in advance.

A:

This is a partial result that I obtained after combining the methods in the following two questions: 
Convergence of $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{\sin(n-m)}{n^2+m^2}$
Does the series $\sum_{n=1}^{\infty}\frac{\sin(\cos(n))}{n}$ converge?
For fixed $n\in\mathbb{N}$, we consider
$$
S_{ n}(N) := \sum_{m=1}^N \sin(\sin(nm)). 
$$
Let $f_n (x) = \sin(\sin(nx))$. This is $2\pi$-periodic function with $\int_0^{2\pi} f_n(x) dx = 0$. Let $\mu\leq 7.6063 $ be the irrationality measure of $\pi$. Then by Koksma's inequality and a bound for the discrepancy, we have
$$
|S_{n}(N)|=O(nN^{1-\frac1{\mu-1}+\epsilon}).
$$
Here, the factor $n$ appears due to the variation of $f_n$. 
By partial summation, we have
$$
\sum_{m=1}^{\infty} \frac{\sin(\sin(nm))}{n^2+m^2}=\sum_{m=1}^{\infty} \frac{S_{ n}(m)-S_n(m-1)}{n^2+m^2}=\sum_{m=1}^{\infty} S_n(m)\left( \frac1{n^2+m^2} - \frac1{n^2+(m+1)^2}\right) 
$$
$$
=O\left(\sum_{m=1}^{\infty} \frac{|S_n(m)|m}{(n^2+m^2)^2} \right)
$$
Thus, we can use $S_n(m)=O(nm^{1-\frac1{\mu-1}+\epsilon})$. Note also that we have a trivial bound $S_n(m)=O(m)$. Now, we compare these bounds and split the sum over $m$ into two parts: $m<n^{\mu-1+\delta}$ and $m\geq n^{\mu-1+\delta}$.
The number $\delta>0$ is chosen so that $1-\frac1{\mu-1}+\frac1{\mu-1+\delta}+\epsilon = 1-\epsilon$. 
The sum over $m<n^{\mu-1+\delta}$ is treated by 
$$
\sum_{m<n^{\mu-1+\delta}} \frac{|S_n(m)|m}{(n^2+m^2)^2} =O\left( \sum_{m<n^{\mu-1+\delta}} \frac {m^2}{(n^2+m^2)^2}\right) =O\left( \frac1n\right)
$$ 
which is not any better than the trivial bound $|\sin(\sin(nm))|\leq 1$. 
For the sum over $m\geq n^{\mu-1+\delta}$, it follows that 
$$
|S_n(m)|=O(m^{1-\epsilon})
$$
We have
$$
\sum_{m\geq n^{\mu-1+\delta}} \frac{|S_n(m)|m}{(n^2+m^2)^2}=O\left( \sum_{m\geq n^{\mu-1+\delta}} \frac{ m^{2-\epsilon}}{(n^2+m^2)^2}\right)=O\left(\frac1{n^{1+\epsilon}}\right)
$$
Therefore, what we obtained is the convergence of 
$$
\sum_{n=1}^{\infty} \sum_{m\geq n^{\mu-1+\delta}} \frac{\sin(\sin(nm))}{n^2+m^2}.
$$
which is clearly not enough for proving anything about the original series.
The main difficulty here is the lack of a nontrivial bound for the range $m< n^{\mu-1+\delta}$. 

