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How can Compute in closed form this double summation :

$$\sum_{k≥1,n≥1}\frac{(-1)^{n+k}}{k(k+1+n)^2}$$

I think here can use harmonic series

Actually I don't have any ideas to approach it

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  • $\begingroup$ Maybe you can try something like this: fix $n$ and look at the sum over $k$; separate even and odd terms; write $\frac{1}{k(k+1+n)^2}$ as something like $\frac{A}{k}+\frac{B}{(k+1+n)}+\frac{Ck+D}{(k+1+n)^2}$ and see if you can compute something. I'm not sure if this works, but it would be the first thing I'd try if I had a pen and a paper to write on. $\endgroup$ – Filippo Sneakerhead May 4 at 19:37
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The idea is to replace the double sum by a double integral which then hopefully can be solved. The hope is justified and we find the following

Result

The closed form is

$$s = \sum_{k≥1,n≥1}\frac{(-1)^{n+k}}{k(k+1+n)^2}\\=\frac{1}{2}\zeta(2) - 2 (1-\log(2))-\frac{1}{8} \zeta(3)\simeq 0.0585043\tag{1}$$

Derivation

The replacement starts with

$$\frac{1}{(n+k+1)^2} = \int_{0}^1 \frac{1}{y} \left(\int_0^y x^{n+k}\,dx \right)\,dy\tag{2}$$

Now we do the double sum under the integral

$$\sum_{n=1}^\infty (\sum_{k=1}^\infty \frac{1}{k} (-x)^{n+k}) =\frac{x \log (x+1)}{x+1}\tag{3}$$

then the $x$-integral

$$\int_0^y \frac{x \log (x+1)}{x+1} \, dx = -y-\frac{1}{2} \log ^2(y+1)+(y+1) \log (y+1)\tag{4}$$

and finally the remaining $y$-integral

$$\int_0^1 \frac{1}{y}(-y-\frac{1}{2} \log ^2(y+1)+(y+1) \log (y+1)) \, dy\tag{5}$$

Observing that

$$\int_0^1 \frac{1}{y} (\log(1+y))^2 = \frac{1}{4} \zeta(3)\tag{6a}$$ $$\int_0^1 \frac{1}{y} \log(1+y) = \frac{1}{2} \zeta(2)\tag{6b}$$ $$\int_0^1 \log(1+y) = -1+\log(4)\tag{6c}$$

$(5)$ gives $(1)$.
Q.E.D.

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