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Calculate the series $$\sum_{j=1}^{\infty }\left( \sum_{n=1}^{\infty }\frac{1}{n^2}\ln \left( 1+\frac{n^2}{j^2} \right)-\frac{1}{j^2}\ln\left( 1+\frac{j^2}{n^2} \right) \right)$$

Note that each term in this series is antisymmetric with respect to the permutation of indices $j$ and $n$. When $j \leftrightarrow n$ is replaced, the first term turns into the second with the opposite sign, and vice versa. Therefore, for each pair $(j, n)$ and $(n, j)$ the sum will be reduced, and, in addition, for $j = n$ each term is zero, since

$$ \frac{1}{n^2} \ln \left( 1 + \frac{n^2}{n^2} \right) - \frac{1}{n^2} \ln \left( 1 + \frac{n^2}{n^2} \right) = 0 $$

And therefore the sum for all such antisymmetric terms over all pairs $(j, n)$ is zero. The sum converges absolutely, since the terms decrease quickly enough. But where did I go wrong? The answer is not zero

Parfig
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Xander Henderson Mar 26 '25 at 18:33

1 Answers1

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An interesting problem :) The answer is indeed $\pi$.

The sum is not absolutely convergent - the order of summation counts, and the value of the sum (the sight) depends on the order of summation.

Using the fact that at any fixed $N$ $$\sum_{j=1}^N\left( \sum_{n=1}^N\frac{1}{n^2}\ln \left( 1+\frac{n^2}{j^2} \right)-\frac{1}{j^2}\ln\left( 1+\frac{j^2}{n^2} \right) \right)=0$$ and that at any fixed $N$ the sums $$\sum_{j=1}^N \sum_{n=N+1}^\infty\frac{1}{n^2}\ln \left( 1+\frac{n^2}{j^2} \right)\,\,\text{and}\,\,\sum_{j=1}^N\sum_{n=N+1}^\infty\frac{1}{j^2}\ln\left( 1+\frac{j^2}{n^2} \right)$$ converge (please, see the addendum), our sum becomes $$S=\sum_{j=1}^{\infty }\left( \sum_{n=1}^{\infty }\frac{1}{n^2}\ln \left( 1+\frac{n^2}{j^2} \right)-\frac{1}{j^2}\ln\left( 1+\frac{j^2}{n^2} \right) \right)$$ $$=\lim_{N\to\infty}\sum_{j=N}^{\infty } \left(\sum_{n=N}^{\infty }\frac{1}{n^2}\ln \left( 1+\frac{n^2}{j^2} \right)-\frac{1}{j^2}\ln\left( 1+\frac{j^2}{n^2} \right)\right) $$ $$=\lim_{N\to\infty}\sum_{j=0}^{\infty } \left(\sum_{n=0}^{\infty }\frac{1}{(N+n)^2}\ln \left( 1+\frac{(N+n)^2}{(N+j)^2} \right)-\frac{1}{(N+j)^2}\ln\left( 1+\frac{(N+j)^2}{(N+n)^2} \right)\right) $$ $$=\lim_{N\to\infty}\frac1{N^2}\sum_{j=0}^{\infty } \left(\sum_{n=0}^{\infty }\frac{1}{(1+\frac n N)^2}\ln \left( 1+\frac{(1+\frac nN)^2}{(1+\frac j N)^2} \right)-\frac{1}{(1+\frac j N)^2}\ln\left( 1+\frac{(1+\frac j N)^2}{(1+\frac n N)^2} \right)\right) $$ $$=\int_1^\infty dy\left(\int_1^\infty dx\left(\frac1{x^2}\ln\Big(1+\frac{x^2}{y^2}\Big)-\frac1{y^2}\ln\Big(1+\frac{y^2}{x^2}\Big)\right)\right)$$ where, integrating by parts, $$\int_1^R dx\left(\frac1{x^2}\ln\Big(1+\frac{x^2}{y^2}\Big)-\frac1{y^2}\ln\Big(1+\frac{y^2}{x^2}\Big)\right)$$ $$=-\frac1x\ln(x^2+y^2)\,\bigg|_{x=1}^{x=R}+2\int_1^R\frac{dx}{x^2+y^2}-2\ln y-\frac x{y^2}\ln(x^2+y^2)\,\bigg|_1^R+\frac2{y^2}\int_1^R\frac{x^2}{x^2+y^2}dx+\frac2{y^2}(x\ln x-x)\,\bigg|_1^R$$ Taking the limit $R\to\infty$ $$\int_1^\infty dx\left(\frac1{x^2}\ln\Big(1+\frac{x^2}{y^2}\Big)-\frac1{y^2}\ln\Big(1+\frac{y^2}{x^2}\Big)\right)=\ln(1+y^2)-2\ln y+\frac{\ln(1+y^2)}{y^2}$$ Integrating with respect to $y$ $$S=\int_1^\infty\left(\ln(1+y^2)-2\ln y+\frac{\ln(1+y^2)}{y^2}\right)dy =\pi$$

$\bf{Addendum}$

In order to skip the pairs $(j,n)$ such that $1<j\leqslant N, \,n>N+1$ and $1<n\leqslant N,\,j>N+1$ it is enough to ensure that both sums $\sum_{j=1}^N \sum_{n=N+1}^\infty\frac{1}{n^2}\ln \left( 1+\frac{n^2}{j^2} \right)\,\,\text{and}\,\,\sum_{j=1}^N\sum_{n=N+1}^\infty\frac{1}{j^2}\ln\left( 1+\frac{j^2}{n^2} \right)$ - converges.

Indeed, $$\sum_{j=1}^N \sum_{n=N+1}^\infty\frac{1}{n^2}\ln \left( 1+\frac{n^2}{j^2} \right)<N\sum_{n=N+1}^\infty\frac{\ln(1+n^2)}{n^2}=F(N)$$ $$\sum_{j=1}^N\sum_{n=N+1}^\infty\frac{1}{j^2}\ln\left( 1+\frac{j^2}{n^2} \right)<\sum_{j=1}^N\frac{1}{j^2}\int_{n=N}^\infty\ln\Big(1+\frac{j^2}{n^2}\Big)dn$$ $$<\sum_{j=1}^N\frac1{j^2}\left(2N\ln N+N\ln(N^2+j^2)+j\arctan\frac jN\right)=G(N)$$ where both functions $F(N),\,G(N)$ are finit at any fixed $N$. It follows that we are allowed to change the order of summation $(n,j)$, what provides mutual cancellation of all terms $1<j\leqslant N, \,n>N$ and $1<n\leqslant N,\,j>N$ of the desired sum $S$

Svyatoslav
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    For fixed $j$ and $n\to\infty$, $\frac1{j^2}\log\bigl(1+\frac{j^2}{n^2}\bigr)\sim\frac1{n^2}$ goes to $0$, not to $1$. – Emil Jeřábek Mar 26 '25 at 17:44
  • Yes, correct. Summation with respect to $n$ gives $\sim\frac{j^2}{j^2}\sum\frac1{n^2}\sim const$ – Svyatoslav Mar 26 '25 at 17:47
  • Well, no. The asymptotics only kicks in for $n\ge j$ or so, which only makes the sum $O(1/j)$. The part with $n\le j$ also contributes only $O(1/j)$, though this is harder to show, hence for fixed $j$, the inner sum is absolutely convergent with value $O(1/j)$. (This is only an upper bound. For all I know, it may cancel even more.) – Emil Jeřábek Mar 26 '25 at 17:51
  • The next problem is, when going from $\sum_{j=1}^\infty\sum_{n=1}^{\infty}$ to $\sum_{j=N}^\infty\sum_{n=N}^\infty$, the pairs $(j,n)$ with $j<N$ and $n<N$ cancel by the first display, but you lost the pairs $(j,n)$ with $j<N\le n$ or $j\ge N>n$. Where did they disappear? You would need some absolute convergence to cancel them. – Emil Jeřábek Mar 26 '25 at 17:54
  • @Emil Jeřábek Just added the part regarding the cancellation of pairs you pointed out. The sign of the desired sum $S$ depends on the order of summation, what confirms the absence of absolute convergence. Though, it would be nice to get a direct prove of it :) – Svyatoslav Mar 26 '25 at 20:45
  • What justifies expressing the limit as an integral, especially inside a conditionally convergent double sum like this? – Emil Jeřábek Mar 26 '25 at 21:21
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    Yes, thank you very much! – Parfig Mar 26 '25 at 21:40
  • @Emil Jeřábek, please see the approach by Sangchul Lee in his answer. The basic idea is consider the difference between the sum and the integral and show that it tends to zero at $N\to\infty$. All details you can find and elaborate using the link - https://math.stackexchange.com/questions/4441546/how-do-i-solve-the-double-summation-sum-n-1-infty-sum-m-1-infty-f/4441629#4441629 Good luck! – Svyatoslav Mar 26 '25 at 21:41
  • Well, since the whole point here is that the sum is nonabsolutely convergent, and therefore various intuitive manipulations are not valid, I’d expect you to include a proper argument in your answer instead of hand-waving. As far as I can see, what you have written here so far is just a heuristic, not a proof, and I see no convincing reason that it should be correct. – Emil Jeřábek Mar 27 '25 at 10:32
  • @Emil Jeřábek, you may like my solution or you may not, but I strongly recommend you to exclude the insulting phrases like "hand waving" when, at least, talking to me. I may also suppose that this phrase is generally inappropriate on this side . Now, to the constructive part. I can justify the limit as an integral, but I won't. I'm not obliged. I gave you a tool of how it can be done - please, share with me where you stopped. Or, better, take the floor and share your own approach to the problem - what, IMHO, will bring more value to the site than the declarative judgments. – Svyatoslav Mar 27 '25 at 14:00
  • You may either fix the serious holes in your argument, or you may shout at the messenger.The choice is yours. – Emil Jeřábek Mar 27 '25 at 14:06
  • @Emil Jeřábek, my choice is to stop communication with you. Please, do me a favor and leave me alone – Svyatoslav Mar 27 '25 at 14:09