Evaluating $\sum_{j\geqslant1}\sum_{k\geqslant1}(-1)^{k+j}\frac{(2k-1)+i(2j-1)}{((2k-1)^{2}+(2j-1)^{2})^{3/2}}.$ After a test I've taken, I considered an infinite grid of eletric charges and wondered the resultant force at the origin. The origin has a charge $+1$ and every gaussian integer $a+bi$ in the first quadrant (only) with odd coefficients has charges too: $+1$ if $a+b\equiv0\pmod4$ and $-1$ if $a+b\equiv2\pmod4$. Every number $z$ applies a force of modulus $1/|z|^2$ at the origin. Working things out, I found the total force at the origin to be $$\sum_{j\geqslant1}\sum_{k\geqslant1}\frac{\left(-1\right)^{k+j}\left(\left(2k-1\right)+i\left(2j-1\right)\right)}{\left(\left(2k-1\right)^{2}+\left(2j-1\right)^{2}\right)^{3/2}}.$$
I don't have access to Mathematica/Maple nor have studied convergence/divergence tests thoroughly. Does it converge? If yes, does it have a nice closed form?
 A: If you type 

sum from j=1 to infinity of sum from k=1 to infinity of [(-1)^(k+j)((2k-1)+i*(2j-1))]/[((2k-1)^2+(2j-1)^2)^(3/2)] 

into Wolfram Alpha, you get that the sum does not converge by the limit test. (Check it and make sure I haven't typed it in there wrong.) Wolfram Alpha is a free tool that can do nice calculations like this for you.
A: This is my justification of why this sum should converge: picking up the first sum and calling $2j-1=a$, we have $$(-1)^j\left(\sum_{k\geqslant 1}\frac{(-1)^k(2k-1)}{((2k-1)^2+a^2)^{3/2}}+ia\sum_{k\geqslant1}\frac{(-1)^k}{((2k-1)^2+a^2)^{3/2}}\right)\\\color{Red}{\leqslant_\star}(-1)^j\left(\frac{1}{2^{3/2}a^{3/2}}\sum_{k\geqslant1}\frac{(-1)^k}{(2k-1)^{1/2}}+\frac{ia}{2^{3/2}a^{3/2}}\sum_{k\geqslant1}\frac{(-1)^k}{(2k-1)^{3/2}}\right)\tag{$\color{Red}{\text{AM-GM}}$}\\=(-1)^j\left(\frac{R+iaS}{2^{3/2}a^{3/2}}\right).$$ $R$ and $S$ both converge by the alternating series test. Now, let's put them in the second sum: $$\text{Original sum}\leqslant_\star\frac{1}{2^{3/2}}\left(R\sum_{j\geqslant1}\frac{(-1)^j}{(2j-1)^{3/2}}+iS\sum_{j\geqslant1}\frac{(-1)^j}{(2j-1)^{1/2}}\right)=\frac{RS}{2^{3/2}}\left(1+i\right).$$
Therefore, the sum converges. I abused the notation using order and complex numbers together, but it can be easily fixed by separating real and imaginary parts. We can see this as $a+bi\leqslant_\star c+ id\iff a\leqslant_\star c \,\text{and}\, b\leqslant_\star d$. 
