TITLE: Prove that $2-\cfrac{\pi^2}{6-\cfrac{\pi^2}{10-\cfrac{\pi^2}{14-\cfrac{\pi^2}{...}}}} = 0$
QUESTION [11 upvotes]: Prove that
$$2 - \cfrac{\pi^2}{6-\cfrac{\pi^2}{10 - \cfrac{\pi^2}{14-\cfrac{\pi^2}{...}}}} = 0$$
My thoughts: One common approahch is setting $x = \text{LHS}$ and we express LHS as $x = \frac{\pi^2}{f(x)}$. But it's hard to translate the denominator as a function of $x$
$\pi^2$ also reminds everybody of the basel problem $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ but what can we do about it?

REPLY [7 votes]: Another answer.
HERE we find the formula
$$
{\displaystyle e^{\frac {x}{y}}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}} }$$
Substitute $x=i\pi, y=1$
$$
{\displaystyle e^{i\pi}=1+{\cfrac {2i\pi}{2 -i\pi+{\cfrac {-\pi^{2}}{6 +{\cfrac {-\pi^{2}}{10 +{\cfrac {-\pi^{2}}{14 +{\cfrac {-\pi^{2}}{18 +\ddots }}}}}}}}}}}
$$
We recognize most of the required terms inside this.  Rearrange this (using $e^{i\pi}=-1$) to get
$$
2 - \cfrac{\pi^2}{6-\cfrac{\pi^2}{10 - \cfrac{\pi^2}{14-\cfrac{\pi^2}{...}}}} = 0
$$