Prove that $2-\cfrac{\pi^2}{6-\cfrac{\pi^2}{10-\cfrac{\pi^2}{14-\cfrac{\pi^2}{...}}}} = 0$ 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?
 A: 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
$$
2 - \cfrac{\pi^2}{6-\cfrac{\pi^2}{10 - \cfrac{\pi^2}{14-\cfrac{\pi^2}{...}}}} = 0
$$
A: You could define it in terms of a sequence:
$$a_n=(4n+2)-\frac{\pi^2}{a_{n+1}}$$
then say:
$$a_n+\frac{\pi^2}{a_{n+1}}=4n+2$$
solving for $a_0, $but the problem is that we do not know any of the values of $a$...

if you start expanding out you get:
$$2=\frac{\pi^2}{...}$$
$$2\left(6-\frac{\pi^2}{...}\right)=\pi^2$$
but I havent made any progress with this
A: The standard reference for continued fractions (at least as of 100 years ago) is
Perron, Oskar, Die Lehre von den Kettenbrüchen. Leipzig—Berlin: B. G. Teubner. xiii, 520 S. $(8^\circ)$ (1913). ZBL43.0283.04.
Section 81, Satz 3 states (in modern language): Consider the continued fraction
$$
b_0 + \frac{a_1}{b_1 + \displaystyle\frac{a_2}{b_2+\ddots}}
$$
with $a_n = a$ and $b_n = dn+c$.  Provided $a \ne 0, c \ne 0,$ and $d \ne 0$, the value of the continued fraction is
$$
V = \frac{c \; {}_0F_1(;c/d;a/d^2)}{\;{}_0F_1(;1+c/d;a/d^2)\;}
$$

Take $a=-\pi^2$ so $a_n = -\pi^2$ and
$c=2, d=4$ so $b_n = 4n+2$.
The value of the continued fraction is computed as
$$
V = \frac{\;2\;{}_0F_1(;1/2;-\pi^2/16)\;}{{}_0F_1(;3/2;-\pi^2/16)}
$$
But the numerator is $2\cos(2\sqrt{\pi^2/16}\;) = 0$ and the denominator is $2/\pi$, so
$$
V = 0 .
$$
