$\nabla \cdot \mathcal{B} = 0$

$\frac{3}{4 \pi} \sqrt{4 \cdot x^2 12}$

$$ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6} e^{i \pi} + 1 = 0 \\ $$

$$ \begin{align*} \int x^2 dx & = \frac{1}{3}x^3 \therefore\quad\int_0^1 x^2 dx &= \frac{1}{3} \end{align*} $$