Basically I'm stuck with this double summation. I want some help evaluating this summation. $$ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{m^2 - n^2}{(m^2 + n^2)^2} $$

Am I allowed to change the order of summation in this?

  • $\begingroup$ Possibly a partial fraction decomposition into poly gamma functions might help, but I’ve not followed the thought through $\endgroup$
    – FShrike
    May 2 at 18:36
  • $\begingroup$ Looks like the series isn't absolutely convergent, so reordering isn't allowed. $\endgroup$ May 2 at 19:12
  • $\begingroup$ Also this has come up before here, although there seems to be some disagreement. $\endgroup$
    – RRL
    May 2 at 19:24
  • $\begingroup$ @RRL A conditionally convergent series can be rearranged to sum to any real number. A reordering isn't allowed unless you can prove it doesn't change the sum, and in this case swapping the order of summation actually does change the sum. $\endgroup$ May 2 at 19:30
  • $\begingroup$ Unless the sum is $0$ in which case both iterated series are equal. The iterated series are not arbitrary rearrangements. But I would agree that without absolute convergence you can rearrange to get other values. In this case the sum appears to be $\pm \frac{\pi}{4}$ depending on the order. I would mark this as a duplicate except the linked question asks about $\sum_{m> n> 0}$. $\endgroup$
    – RRL
    May 2 at 19:33

2 Answers 2


Let me give my 5 cents and present one more (heuristic) solution based on the Euler-Maclaurin summation formula. First, we note that for any finite $k\quad \sum_{n=1}^k \sum_{m=1}^k\frac{m^2 - n^2}{(m^2 + n^2)^2}=0$. Therefore, we can consider $$S=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{m^2 - n^2}{(m^2 + n^2)^2}=\lim_{k\to\infty}\sum_{n=k}^\infty \sum_{m=k}^\infty\frac{m^2 - n^2}{(m^2 + n^2)^2}=\lim_{k\to\infty}S(k)$$ To evaluate $S(k)$ we can use Euler-Maclaurin summation formula which perfectly works in this case: $$S(k)=\sum_{n=k}^\infty\bigg(\int_k^\infty \frac{m^2 - n^2}{(m^2 + n^2)^2}dm+\frac{1}{2}\Big(\frac{m^2 - n^2}{(m^2 + n^2)^2}\,\bigg|^{m=k}+\frac{m^2 - n^2}{(m^2 + n^2)^2}\,\bigg|^{m=\infty}\,\Big)+...\bigg)$$ Other terms contain higher derivatives with respect to $m$ and, therefore, higher power of $\frac{1}{k}$ $$S(k)\sim \sum_{n=k}^\infty\bigg(\int_k^\infty \frac{m^2 - n^2}{(m^2 + n^2)^2}dm+\frac{1}{2}\frac{k^2 - n^2}{(k^2 + n^2)^2}\bigg)$$ Using the Euler-Maclaurin formula with respect to $n$, we can treat the second term as $$\frac{1}{2}\sum_{n=k}^\infty\frac{k^2 - n^2}{(k^2 + n^2)^2}=\frac{1}{2}\int_k^\infty\frac{k^2 - n^2}{(k^2 + n^2)^2}dn+O\Big(\frac{1}{k^2}\Big)$$ $$=\frac{1}{2k}\int_1^\infty\frac{dx}{1+x^2}-\frac{1}{k}\int_1^\infty\frac{x^2\,dx}{(1+x^2)^2}+O\Big(\frac{1}{k^2}\Big)=O\Big(\frac{1}{k}\Big)$$ Therefore, all contribution comes from the term $$S(k)= \sum_{n=k}^\infty\int_k^\infty \frac{m^2 - n^2}{(m^2 + n^2)^2}dm+O\Big(\frac{1}{k}\Big)=\int_k^\infty dn\int_k^\infty \frac{m^2 - n^2}{(m^2 + n^2)^2}dm+O\Big(\frac{1}{k}\Big)$$ $$=\int_1^\infty dx\int_1^\infty\frac{y^2-x^2}{(y^2+x^2)^2}ds+O\Big(\frac{1}{k}\Big)=\int_1^\infty \frac{dx}{x}\int_{1/x}^\infty\frac{s^2-1}{(s^2+1)^2}dx+O\Big(\frac{1}{k}\Big)$$ Integration with respect to $s$ is straightforward: $$S(k)=\int_1^\infty \frac{dx}{x}\int_{1/x}^\infty\Big(\frac{1}{s^2+1}-\frac{2}{(s^2+1)^2}\Big)ds+O\Big(\frac{1}{k}\Big)$$ $$=\int_1^\infty \frac{dx}{x}\Big(\arctan x-\arctan x+x-\frac{x^3}{1+x^2}\Big)+O\Big(\frac{1}{k}\Big)$$ $$=\int_1^\infty\frac{dx}{1+x^2}+O\Big(\frac{1}{k}\Big)=\frac{\pi}{4}+O\Big(\frac{1}{k}\Big)$$ $$S=\lim_{k\to\infty}S(k)=\frac{\pi}{4}$$

  • 1
    $\begingroup$ Thank you for your beautiful solution sir! $\endgroup$ May 3 at 19:43
  • 2
    $\begingroup$ @Vaibhav C M I'm glad if this helps. Please look at the elegant and professional solutions in the second post. $\endgroup$
    – Svyatoslav
    May 3 at 20:29

Since the sum does not converge absolutely, we are not allowed to rearrange the order of summation. In this answer, we will show that the sum

$$ S := \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{m^2 - n^2}{(m^2 + n^2)^2} $$

converges and find its value.

New Anwer. We will follow @Svyatoslav's approach with some simplification. As in his answer, we note that

$$ S = \lim_{K \to\infty} \sum_{n > K} \sum_{m > K} \frac{m^2 - n^2}{(m^2 + n^2)^2}. $$

To estimate the inner sum, we observe:

Observation. Let $K$ be a positive integer, and let $f : [K, \infty) \to [0, \infty)$ be decreasing and integrable. Then by the triangle inequality, \begin{align*} \left| \int_{K}^{\infty} f(x) \, \mathrm{d}x - \sum_{n > K}^{\infty} f(n) \right| &\leq \sum_{n > K}^{\infty} \left| \int_{n-1}^{n} f(x) \, \mathrm{d}x - f(n) \right| \\ &\leq \sum_{n > K}^{\infty} \bigl[ f(n-1) - f(n) \bigr] \\ &\leq f(K). \end{align*}

So, by noting that both $x \mapsto \frac{1}{x^2 + n^2}$ and $x \mapsto \frac{2n^2}{(x^2+n^2)^2}$ satisfy the hypotheses of the lemma, we get

$$ \left| \sum_{m > K} \frac{m^2 - n^2}{(m^2 + n^2)^2} - \int_{K}^{\infty} \frac{x^2 - n^2}{(x^2 + n^2)^2} \, \mathrm{d}x \right| \leq \frac{1}{K^2 + n^2} + \frac{2n^2}{(K^2 + n^2)^2} \leq \frac{3}{n^2}. $$


\begin{align*} \sum_{n > K} \sum_{m > K} \frac{m^2 - n^2}{(m^2 + n^2)^2} &= \sum_{n > K} \biggl[ \int_{K}^{\infty} \frac{x^2 - n^2}{(x^2 + n^2)^2} \, \mathrm{d}x + \mathcal{O}\biggl(\frac{1}{n^2}\biggr) \biggr] \\ &= \sum_{n > K} \biggl[ \frac{K}{K^2 + n^2} + \mathcal{O}\biggl(\frac{1}{n^2}\biggr) \biggr] \\ &= \sum_{n > K} \frac{1}{1 + (n/K)^2} \frac{1}{K} + \mathcal{O}\biggl(\frac{1}{K}\biggr) \end{align*}

Letting $K \to \infty$, this converges to

$$ S = \int_{1}^{\infty} \frac{1}{1+x^2} \, \mathrm{d}x = \frac{\pi}{4}. $$

Old Answer. First, using the identity

$$ \int_{0}^{\infty} x \cos(nx) e^{-mx} \, \mathrm{d}x = \frac{m^2 - n^2}{(m^2 + n^2)^2}, \qquad m, n > 0,$$

and Fubini's theorem, we obtain

\begin{align*} S(n) := \sum_{m=1}^{\infty} \frac{m^2 - n^2}{(m^2 + n^2)^2} &= \sum_{m=1}^{\infty} \int_{0}^{\infty} x \cos(nx) e^{-mx} \, \mathrm{d}x \\ &= \int_{0}^{\infty} \sum_{m=1}^{\infty} x \cos(nx) e^{-mx} \, \mathrm{d}x \\ &= \int_{0}^{\infty} \frac{x \cos(nx)}{e^x - 1} \, \mathrm{d}x = \int_{0}^{2\pi} f(x) \cos(nx) \, \mathrm{d}x, \end{align*}

where $f$ is the function defined by

$$ f(x) = \sum_{k=0}^{\infty} \frac{x+2k\pi}{e^{x+2k\pi} - 1}. $$

Now consider the $2\pi$-periodic modification $\tilde{f}(x) = f(x \text{ mod } 2\pi)$. Using the general theory of Fourier series, we can check that

$$ \frac{a_0}{2} + \sum_{n=1}^{\infty} \bigl[ a_n \cos(n\theta) + b_n \sin(n\theta) \bigr] = \frac{\tilde{f}(\theta^+) + \tilde{f}(\theta^-)}{2} \tag{*} $$


$$ a_n = \frac{S(n)}{\pi} = \frac{1}{\pi} \int_{0}^{2\pi} f(x)\cos(nx) \, \mathrm{d}x \qquad\text{and}\qquad b_n = \frac{1}{\pi} \int_{0}^{2\pi} f(x)\sin(nx) \, \mathrm{d}x. $$

Plugging $\theta = 0$ into $\text{(*)}$ and comparing this with $S = \sum_{n=1}^{\infty} S(n)$, it follows that

\begin{align*} S = \sum_{n=1}^{\infty} S(n) &= \frac{\pi}{2}[f(0) + f(2\pi)] - \frac{1}{2} \int_{0}^{2\pi} f(x) \, \mathrm{d}x \\ &= \frac{\pi}{2} + 2\pi^2 \sum_{k=1}^{\infty} \frac{k}{e^{2k\pi} - 1} - \frac{1}{2} \int_{0}^{\infty} \frac{x}{e^x - 1} \, \mathrm{d}x \end{align*}

The sum and the integral in the last line are well known:

$$ \sum_{k=1}^{\infty} \frac{k}{e^{2k\pi} - 1} = \frac{1}{24} - \frac{1}{8\pi} \qquad\text{and}\qquad \int_{0}^{\infty} \frac{x}{e^x - 1} \, \mathrm{d}x = \zeta(2) = \frac{\pi^2}{6}. $$

see this posting for instance. Therefore

$$ S = \frac{\pi}{4}. $$

  • 2
    $\begingroup$ Nice solution! An amateur always looks at the work of a professional with fascination :) An elegant proof of $\sum_{n=1}^{\infty}\frac{1}{\sinh^2\!\pi n}=\frac{1}{6}-\frac{1}{2\pi}$ using complex integration is also here math.stackexchange.com/questions/1866502/… (a third post) $\endgroup$
    – Svyatoslav
    May 3 at 4:57
  • 2
    $\begingroup$ @Svyatoslav, Thank you! I also like your answer very much! I think we can simplify your approach a bit, so I will update my answer accordingly. $\endgroup$ May 3 at 10:26
  • 1
    $\begingroup$ Thank you for your simplification! It this form it is much easier, just three lines :) $\endgroup$
    – Svyatoslav
    May 3 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.