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As in this question, I am trying to understand convergence behavior of a double sum by numerical investigation. The sum is $$ \sum_{j=1}^\infty\sum_{k=1, k\neq j}^\infty \frac{1}{j^2-k^2}$$ Due to symmetry and cancellation, I get the result $$\lim_{J \to \infty}\sum_{j=1}^J\sum_{k=1,k \neq j}^J \frac{1}{j^2-k^2}= 0$$

But if I evaluate $\sum_{j=1}^J\sum_{k=1, k \neq j}^K \frac{1}{j^2-k^2}$ and increase $J$ and $K$ with $J \neq K$ there seems to be convergence to different values. For example, if $K = 10 J$ then it appears to converge to $\approx 1.133$ and if $K = J^2$ then $\approx 1.216$.

What is happening here?

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  • $\begingroup$ I'm not certain what you're asking here. Were these results unexpected for some reason? $\endgroup$ Oct 2 '18 at 22:33
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    $\begingroup$ There is nothing strange. A similar situation occurs for $$\sum_{m\geq 1}\sum_{n\geq 1}\frac{m^2-n^2}{(m^2+n^2)^2}\neq\sum_{n\geq 1}\sum_{m\geq 1}\frac{m^2-n^2}{(m^2+n^2)^2}.$$ We do not have absolute convergence since every number which is not of the form $4k+2$ can be written as a difference of two squares, hence $$\sum_{\substack{j,k\geq 1\\ j\neq k}}\frac{1}{|j^2-k^2|}\geq \sum_{n\geq 0}\left(\frac{1}{4n+1}+\frac{1}{4n+3}+\frac{1}{4n+4}\right)=+\infty.$$ $\endgroup$ Oct 2 '18 at 22:36
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It is not surprising that your numerical results are confusing.

First, be aware there is a distinction between the double series and an iterated series. When both sides converge it is true that

$$\sum_{j,k=1}^\infty a_{jk} = \sum_{j=1}^\infty \sum_{k=1}^\infty a_{jk}$$

where convergence of the double series on the LHS means $\left|\sum_{j=1}^J\sum_{k=1}^Ka_{jk} - S\right| < \epsilon$ when $J,K > N(\epsilon)$ and convergence of the iterated series on the RHS means $S = \lim_{J \to \infty} \left(\lim_{K \to \infty}\sum_{j=1}^J\sum_{k=1}^Ka_{jk}\right) = S.$

When the terms can change sign, the double series can fail to converge even if iterated series converge.

In this case, the double series is not absolutely convergent since

$$\sum_{j=1}^J\sum_{k=1\\j \neq k}^K \frac1{|k^2-j^2|} > \sum_{j=1}^J\sum_{k=1\\j \neq k}^K \frac1{|k^2+j^2|} , $$

and the RHS diverges by the integral test.

The series is also not conditionally convergent since it can be shown that for the iterated series,

$$\tag{*}\sum_{j=1}^\infty \sum_{k=1\\j\neq k}^\infty \frac{1}{k^2 - j^2} = -\sum_{k=1}^\infty \sum_{j=1\\j \neq k}^\infty \frac{1}{k^2-j^2} = \frac{\pi^2}{8} \approx 1.2337$$

Taking $K = J^2$, your computations appear to be approaching this value.

When the iterated series have opposite signs, the double series can be convergent only if the value is $0$, which is not the case. The result (*) can be obtained using

$$\sum_{k=1\\j\neq k}^K \frac{1}{k^2-j^2} = \frac{1}{2j}\sum_{k=1\\j\neq k}^K \left(\frac{1}{k-j} - \frac{1}{k+j} \right) \\= \frac{1}{2j}\left(-\sum_{k=1}^{j-1}\frac{1}{k} + \sum_{k=1}^{K-j}\frac{1}{k} - \sum_{k=j+1}^{K+j}\frac{1}{k} + \frac{1}{2j} \right)\\= \frac{1}{2j}\left(\frac{1}{j} +\sum_{k=j+1}^{K-j}\frac{1}{k} - \sum_{k=j+1}^{K+j}\frac{1}{k} + \frac{1}{2j} \right)\\ = \frac{1}{2j}\left(\frac{3}{2j} - \sum_{k=-j+1}^{j} \frac{1}{K+k}\right) $$

and,

$$\sum_{k=1\\j\neq k}^\infty \frac{1}{k^2-j^2} = \lim_{K \to \infty}\frac{1}{2j}\left(\frac{3}{2j} - \sum_{k=-n+1}^{n} \frac{1}{K+k}\right) = \frac{3}{4j^2}$$

Thus,

$$\sum_{j=1}^\infty \sum_{k=1\\j\neq k}^\infty \frac{1}{k^2 - j^2} = \sum_{j=1}^\infty \frac{3}{4j^2} = \frac{3}{4} \frac{\pi^2}{6}$$

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    $\begingroup$ Thank you for this. Could you please explain how you get the sum in the form $\frac{3}{2j} - \sum_{k=-n+1}^{n} \frac{1}{K+k}$ $\endgroup$
    – WoodWorker
    Oct 2 '18 at 23:36
  • $\begingroup$ @WoodWorker: I'll add more steps in answer. $\endgroup$
    – RRL
    Oct 3 '18 at 2:52

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