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How to compute following double series,

$$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty} \cfrac{(-1)^{(i+j)}}{i^2+2nij+j^2} $$

where $n\in \mathbb{N}$.

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It looks almost like an Epstein zeta function... –  J. M. Jul 11 '12 at 17:58

2 Answers 2

This double series is not absolutely convergent. You might try a kind of Abel summation: $\lim_{z \to -1+} f_n(z)$ where $$f_n(z) = \sum_{i=1}^\infty \sum_{j=1}^\infty \dfrac{z^{i+j}}{i^2 + 2nij + j^2}$$ But I would find a closed form surprising.

It might be interesting to try this asymptotically as $n \to \infty$, using $$ \dfrac{1}{i^2 + 2nij + j^2} = {\frac {1}{2ijn}}-{\frac {{i}^{2}+{j}^{2}}{4{i}^{2}{j}^{2}{n}^ {2}}}+{\frac { ( {i}^{2}+{j}^{2} ) ^{2}}{8 {i}^{3}{j}^{3 }{n}^{3}}} - \dfrac{(i^2 + j^2)^3}{16 i^4 j^4 n^4}+\ldots$$ With Maple's help, I get $$ f_n(z) \sim \dfrac{(\ln(1-z))^2}{2n} - \dfrac{z \ \text{polylog}(2,z)}{2 (1-z) n^2} + \dfrac{z \ \text{polylog}(3,z)}{4 (z-1)^2 n^3} + \dfrac{(\ln(1-z))^2}{4 n^3} + + \dfrac{z(z+1) \ \text{polylog}(4,z)}{8 (z-1)^3 n^4 } + \dfrac{3 z \ \text{polylog}(2,z)}{8 (z-1) n^4} + \ldots$$

which as $z \to -1+$ becomes

$$ \dfrac{(\ln 2)^2}{2n} - \dfrac{\pi^2}{48 n^2} + \dfrac{3 \zeta(3) + 16 (\ln 2)^2}{64 n^3} - \dfrac{\pi^2}{64 n^4} + \ldots$$

There must be a pattern here...

EDIT: The coefficient of $n^{-m}$ is $$ \eqalign{\sum_{i=1}^\infty \sum_{j=1}^\infty &\dfrac{(-1)^{m+1}}{2^m i^m j^m} (i^2 + j^2)^{m-1} z^{i+j} = \dfrac{(-1)^{m+1}}{2^m} \sum_{k=0}^{m-1} {{m-1} \choose k} \sum_{i=1}^\infty i^{2k-m} z^i \sum_{j=1}^\infty j^{m-2-2k} z^j\cr &= \dfrac{(-1)^{m+1}}{2^m} \sum_{k=0}^{m-1} {{m-1} \choose k} \operatorname{polylog}(-2k+m,z) \operatorname{polylog}(-m+2+2k,z)\cr}$$ Note that for nonnegative integers $p$, $ \operatorname{polylog}(-p,z) = z P_p(1-z)/(1-z)^{p+1}$ where $P_p$ is a polynomial of degree $p-1$. If $m$ is even, each term involves a polylog of positive index times a rational function. If $m$ is odd, there is also (for $k=(m-1)/2$) one term $2^{-m} {{m-1} \choose {(m-1)/2}} \ln(1-z)^2$ since $\operatorname{polylog}(1,z) = -\log(1-z)$.

When we take $z \to -1$, $\operatorname{polylog}(p,-1) = (2^{1-p}-1) \zeta(p)$ for integers $p \ge 2$, while $\operatorname{polylog}(-p,-1) = -(2^{p+1}-1) B_{p+1}/(p+1)$, where $B_{p+1}$ is the $(p+1)$'th Bernoulli number (which is $0$ for even $p\ge 2$). Thus for even $m$, the only terms that survive are those for $k=m/2$ and $m/2-1$, and the coefficient of $n^{-m}$ is $\displaystyle - 2^{-m} {{m-1} \choose {m/2}} \dfrac{\pi^2}{12}$. For odd $m$, we get $$ 2^{-m} {{m-1} \choose {(m-1)/2}} (\ln 2)^2 + 2^{2-2m} \sum_{k=0}^{(m-3)/2} {{m-1}\choose k} (2^{m-2k-1}-1)^2 2^{2k} \dfrac{B_{m-2k-1}}{m-2k-1} \zeta(m-2k)$$

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I believe there's a sign error in your last expression. Otherwise, very nice. (+1) –  user26872 Jul 11 '12 at 7:11
    
Fixed the sign error, thanks. –  Robert Israel Jul 11 '12 at 16:02

When $n=1$, you get $${1\over2^2}-{2\over3^2}+{3\over4^2}-{4\over5^2}+\cdots={\pi^2\over12}-\log2,$$ series number 347 in Jolley, Summation of Series. Do you have a reason to think there's a closed form for general $n$?

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even for $n=1$ I would be careful: the double sum is not absolutely convergent, so reordering of terms might change it. –  user8268 Jul 11 '12 at 6:28

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