# 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.

• So, what is the 5 in the title? – Gerry Myerson Jan 4 '14 at 16:26
• @GerryMyerson it is very large and fat, so must be important... – Igor Rivin Jan 4 '14 at 16:32
• @Igor, it was smaller and skinnier when I asked, and there was an equals sign in front of it. It's more mysterious now than it was then. Let's see what happens on the next edit. – Gerry Myerson Jan 4 '14 at 16:35
• I dont understand the irony, seriously... – user113600 Jan 4 '14 at 17:32
• Because it is not absolutely convergent, the exact order in which you sum it matters. It can be made to converge to whatever you want. No unique sum. – Somos Jun 5 '17 at 12:55

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 $m< n^{\mu-1+\delta}$.

• I see that you want to find the inner summation first and then do the outer summation. That specifies the order of summation. I did not see that before. – Somos Jun 6 '17 at 11:45
• It seems that $\mu=2$ is still not enough to prove convergence/divergence. – Sungjin Kim Jun 6 '17 at 21:36