I would like to calculate the $j$-invariance of tori $\Sigma=\mathbb{C}/L$, where lattice $L:=<1,i>$. And I try to show that the ellipict $y^2=x^3-x$ is isomorphism with $\mathbb{C}/L$. From the Weierstrass function $\mathscr{P}$, we have $\forall \tau \in \mathbb{H}$ $$g_{2}(\tau)=60G_{4}(\tau), g_{3}(\tau)=140G_{6}(\tau),$$ Here, we know $$G_{4}(\tau)=G_{4}(L)=\sum_{(m,n)\in \mathbb{Z}^2 \setminus (0,0)} \frac{1}{(m+ni)^4},$$ $$G_{6}(\tau)=\sum_{(m,n)\in \mathbb{Z}^2 \setminus (0,0)} \frac{1}{(m+ni)^6}$$
Is there an answer to two series?
EDIT:
I look up a theorem: If $k \geq 4$ is even, and $Im(\tau)>0$, then $$E_{k}(\tau)=2\zeta(k)+\frac{2(-1)^{\frac{k}{2}}}{(k-1)!}\sum_{r=1}^{\infty}\sigma_{k-1}(r)e^{2\pi i \tau r}，$$ where divisor function $\sigma_{l}=\sum_{d|r}d^l.$
But how to get the answer $j$-invariance of tori: $$ j=\frac{g_{2}^3(\tau)}{\Delta},$$ where $\Delta(\tau)=g_{2}^3(\tau)-27g_{3}^2(\tau)$.
Consider the Weierstrass function $\wp(z)$ with $g_2 = 1, g_3 = 0$. $\omega_1, \omega_2$ be its periods. The discriminant is positive, hence one period can be taken to be real, another purely imaginary. Let $${e_1} = \wp (\frac{{{\omega _1}}}{2}) \qquad {e_2} = \wp (\frac{{{\omega _2}}}{2}) \qquad {e_3} = \wp (\frac{{{\omega _1} + {\omega _2}}}{2})$$ $e_i$ are roots of equation $4x^3-x = 0$. From $\wp'(z)^2 = 4\wp(z)^3 - \wp(z) $, we have $$\frac{{{\omega _1}}}{2} = \int_{{e_1}}^\infty {\frac{1}{{\sqrt {4{x^3} - x} }}dx} = \int_{1/2}^\infty {\frac{1}{{\sqrt {4{x^3} - x} }}dx} = \frac{1}{{4\sqrt \pi }}{\Gamma ^2}\left( {\frac{1}{4}} \right)$$ $$\frac{{{\omega _2}}}{2} = i\int_{ - \infty }^{ - 1/2} {\frac{1}{{\sqrt {x - 4{x^3}} }}dx} = \frac{i}{{4\sqrt \pi }}{\Gamma ^2}\left( {\frac{1}{4}} \right)$$ which is a straightforward calculation using beta function. Hence $${\omega _2} = i{\omega _1},{\rm{ }}{\omega _1} = \frac{1}{{2\sqrt \pi }}{\Gamma ^2}\left( {\frac{1}{4}} \right)$$ So $${G_4} = \frac{{{g_2}}}{{60}} = \frac{1}{{60}} = \sum\limits_{(m,n) \ne (0,0)} {\frac{1}{{{{({\omega _1}m + {\omega _1}in)}^4}}}} $$ $$\sum\limits_{(m,n) \ne (0,0)} {\frac{1}{{{{(m + in)}^4}}}} = \frac{{{\omega ^4}}}{{60}} = \color{blue}{\frac{1}{{960{\pi ^2}}}{\Gamma ^8}\left( {\frac{1}{4}} \right)}$$ Simiarly, $G_6$ when $\tau = i $ is $0$, which can also be shown directly.