Ramanujan – $\pi$, $e$ in $\phi$

$$ {\sqrt{5+\sqrt{5}\over 2}-\phi} = \cfrac{e^{-\frac{2\pi}{5}}}{1 + \cfrac{e^{-2\pi}}{1 + \cfrac{e^{-4\pi}}{1 + \cfrac{e^{-6\pi}} {1 + \cfrac{e^{-8\pi}} {1+\ddots}}}}} $$