print · login   

$ \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}{1+\frac{e^{-8\pi}} {1+\cdots} } } } $

$\displaystyle\sum_{i=1}^{k+1}i$