\displaystyle \begin{aligned} \mathrm{e} &= \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) ^n \\ &\approx 2.71828\,18284\,59045\,23536\,02874\,71352\,\cdots \end{aligned}

\displaystyle \begin{aligned} \sum_{i=1}^n i^3 = \left( \sum_{i=1}^n i \right)^2 \end{aligned}