Problem of the Week #180 - November 10, 2015

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Here is this week's POTW:

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Prove or disprove that the series

$$\sum_{\substack{(m,n)\in \Bbb N^2\\ m\neq n}}\frac{\cos(mn x)}{m^2 - n^2}$$

converges for all $x \in [-1,1]$.
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Congratulations to Opalg for solving this problem correctly! Here is his solution, which also underlines the subtleties of this problem:

Spoiler

The term given by $(m,n)$ is the negative of the term given by $(n,m)$. So each such pair of terms cancels, and if the series converges to anything it must converge to $0$.

But does the series in fact converge at all? The set $\{(m,n) \in \Bbb N^2: m\ne n\}$ is not equipped with a natural total order. So it only makes sense to talk about the series converging if it converges unconditionally (that is, independently of the order of the terms). That in turn is equivalent to absolute convergence. So we should look to see if the subseries of positive terms converges. It is easiest to do that for the point $x=0$. In that case, $\frac{\cos(mn x)}{m^2 - n^2} = \frac{1}{m^2 - n^2}$, and the positive terms are those for which $m>n$. So we need to look at the series $$\sum_{\substack{(m,n)\in \Bbb N^2\\ m> n}}\frac{\cos(mn x)}{m^2 - n^2}.$$ Suppose that we restrict to those terms for which $n = m-1$. Then we get the subseries $$\sum_{m=1}^\infty \frac1{m^2 - (m-1)^2} = \sum_{m=1}^\infty \frac1{2m-1}.$$ That is a variant of the harmonic series, and it diverges. Therefore that whole series of positive terms diverges.Conclusion: the series cannot be said to converge when $x=0$ and therefore it is not the case that it converges for all $x\in [-1,1].$