<s>[INST] Continued fraction for $\sqrt 2$ I discovered this continued fraction for $\sqrt 2$ but could not find any sources in which it appeared. It goes as follows:
\begin{equation}
\sqrt{2} = 1 + \frac{1 + \frac{1 + \frac{1 + \dots}{3 + \dots}}{3 + \frac{1 + \dots}{3 + \dots}}}{3 + \frac{1 + \frac{1 + \dots}{3 + \dots}}{3 + \frac{1 + \dots}{3 + \dots}}}
\end{equation}
I can show that it approaches $\sqrt{2}$ via a computer program, but see no feasible way to prove it.
 [/INST] Let $x = \frac {1 + \frac {1 + \cdots}{3+\cdots}}{3 + \frac {1 + \cdots}{3+\cdots}}$
$x = \frac {1+x}{3+x}\\
x^2+ 3x = 1+x\\
x^2 + 2x - 1 = 0\\
x = -1 \pm \sqrt {2}$
We know that $x>0$ so we can discard the negative value.
$1 + x = \sqrt 2$
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