# A closed form of the double sum $\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2}$

I want to evaluate the double sum $\displaystyle \sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2}$ where I know that the value is $-\frac{17\pi^4}{1440}$ , I have failed to evaluate to the inner finite sum cleverly so as to transform it into a reasonable infinite series.

• Maybe show what you did, we may help you better :) – ParaH2 Aug 14 '16 at 19:00
• Hint: sort by the value of $k=m^2-n^2$. Given such a $k$, the number of times $k$ appears in the sum is the number of factorizations of the type $k=(m+n)(m-n)$, that is, the number of factorizations of $k$ into two factors of the same parity. That should transform the problem into one involving values of (cousins of) the zeta function. – Greg Martin Aug 14 '16 at 19:09
• I understand what you are aiming at, had it been $m^2+n^2$ , the sum of squares function would have been useful then transforming it into a analytical NT problem. – Aditya Narayan Sharma Aug 14 '16 at 19:13
• My first thought would be to factorize the denominator into $(m-n)^2(m+n)^2$, but then it's not entirely clear (to me) how to proceed from there. – Clement C. Aug 14 '16 at 19:20
• Possibly related: math.stackexchange.com/questions/1294185/… – nospoon Aug 14 '16 at 19:24

Hint. One may observe that, for $m^2\neq n^2$, we have $$\frac1{(m^2-n^2)^2}=\int_0^1\int_0^1x^{m+n-1}y^{m-n-1}\log x \log y\:dxdy \tag1$$ giving, by using absolute convergence, $$\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2}=\int_0^1\!\!\int_0^1\frac{-\log x \log y}{(1-x^2)(1+xy)}\:dxdy. \tag2$$ Then, by partial fraction decomposition, one may write

\begin{align} &\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2} \\\\&=\int_0^1\!\!\int_0^1\frac{-\log x \log y}{(1-x^2)(1+xy)}\:dxdy \\\\&=\int_0^1\!\!\int_0^1\left(\frac{-\log x \log y}{2 (1+x) (1-y)}+\frac{-\log x \log y}{2 (1-x)(1+y)}+\frac{y^2\log x \log y}{(1-y^2) (1+x y)}\right)\!dxdy \tag3 \\\\&=-2\cdot \frac{\pi^4}{144}+\frac{\pi^4}{480} \\\\&=-\frac{17\pi^4}{1440} \end{align} as announced.

• What a beautiful answer! – Omar Nagib Aug 14 '16 at 22:30
• @Omar Nagib Thank you very much. – Olivier Oloa Aug 15 '16 at 0:33
• Amazing stuff !!!! , Thanks ! – Aditya Narayan Sharma Aug 15 '16 at 5:40
• @OlivierOloa: Nice and concise! (+1) – Markus Scheuer Aug 15 '16 at 6:36

Let $k\in\mathbb{N}$. The function $\tau:\mathbb{N}\to\mathbb{N}$ is the divisor-counting function. For an odd value of $k$, there are $\left\lceil\frac{1}{2}\tau(k)\right\rceil$ pairs $(m,n)$ with $m\in\mathbb{N}$ and $n\in\mathbb{N}_0$ such that $m>n$ and $k=m^2-n^2$, whereas there are $\left\lceil\frac{1}{2}\tau\left(\frac{k}{4}\right)\right\rceil$ such pairs if $k$ is divisible by $4$. If $k$ is even but not divisible by $4$, then there are no such pairs $(m,n)$.

The required sum is then $$S:=\sum_{m=1}^\infty\,\sum_{n=0}^{m-1}\,\frac{(-1)^{m-n}}{\left(m^2-n^2\right)^2}=\sum_{k\equiv0\!\pmod{4}}\,\frac{\left\lceil\frac{1}{2}\tau\left(\frac{k}{4}\right)\right\rceil}{k^2}-\sum_{k\text{ odd}}\,\frac{\left\lceil\frac{1}{2}\tau(k)\right\rceil}{k^2}\,.$$ Now, by handling perfect-square values of $k$, we see that $$S=\frac{1}{2}\,\sum_{k\equiv0\!\pmod{4}}\,\frac{\tau(k/4)}{k^2}-\frac{1}{2}\sum_{k\text{ odd}}\,\frac{\tau(k)}{k^2}+\frac{1}{2}\,\sum_{j=1}^\infty\,\frac{(-1)^j}{j^4}\,.$$ Hence, with $j:=\frac{k}{4}$ in the first sum, we have $$S=\frac{1}{32}\,\sum_{j=1}^\infty\,\frac{\tau(j)}{j^2}-\frac{1}{2}\,\sum_{k\text{ odd}}\,\frac{\tau(k)}{k^2}-\frac{1}{2}\,\eta(4)\,,$$ where $\eta$ is the Dirichlet eta function. Note that $$\eta(4)=\left(1-\frac{1}{2^3}\right)\,\zeta(4)=\frac{7}{8}\left(\frac{\pi^4}{90}\right)=\frac{7\pi^4}{720}\,,$$ where $\zeta$ is the Riemann zeta function.

Notice that $$\sum_{j=1}^\infty\,\frac{\tau(j)}{j^2}=\prod_{p\text{ prime}}\,\sum_{r=0}^\infty\,\frac{r+1}{p^{2r}}=\prod_{p\text{ prime}}\,\frac{p^4}{\left(p^2-1\right)^2}$$ and that $$\sum_{k\text{ odd}}\,\frac{\tau(k)}{k^2}=\prod_{p\text{ odd prime}}\,\sum_{r=0}^\infty\,\frac{r+1}{p^{2r}}=\prod_{p\text{ odd prime}}\,\frac{p^4}{\left(p^2-1\right)^2}\,,$$ whence $$\sum_{k\text{ odd}}\,\frac{\tau(k)}{k^2}=\frac{9}{16}\,\sum_{j=1}^\infty\,\frac{\tau(j)}{j^2}\,.$$ Now, observe that $$\sum_{j=1}^\infty\,\frac{\tau(j)}{j^2}=\left(\prod_{p\text{ prime}}\,\frac{1}{1-p^{-2}}\right)^2=\big(\zeta(2)\big)^2=\left(\frac{\pi^2}{6}\right)=\frac{\pi^4}{36}\,.$$ Combining all the pieces, we have $$S=\frac{1}{32}\left(\frac{\pi^4}{36}\right)-\frac{1}{2}\left(\frac{9}{16}\right)\left(\frac{\pi^4}{36}\right)-\frac{1}{2}\left(\frac{7\pi^4}{720}\right)=-\frac{17\pi^4}{1440}\,.$$

P.S. We can show that $$\sum_{j=1}^\infty\,\frac{\tau(j)}{j^s}=\big(\zeta(s)\big)^2$$ for all $s\in\mathbb{C}$ with $\text{Re}(s)>1$.

• (+1) So this is the NT method for this sum, Nice observation and thanks – Aditya Narayan Sharma Aug 15 '16 at 5:41