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.
 A: Your continued fraction is essentially the 'infinite image' of the map $x\mapsto f(x)=1+\dfrac x{2+x}$ (can you see why?) and so any limit $\alpha$ of that map must satisfy $\alpha=1+\dfrac{\alpha}{2+\alpha}$. Multiplying both sides by $2+\alpha$ and simplifying gives the equation $\alpha^2=2$, so $\sqrt{2}$ must be the limit if it exists. To show that the limit exists, you want to show that $f(x)$ is what's called a Contraction Mapping over a suitable section of the real line — that is to say, that the distance between $f(x)$ and $f(y)$ is always smaller than the distance between $x$ and $y$ within that domain. The Banach fixed-point theorem then guarantees that it has a (unique) fixed point in that region. The easiest way to show that $f()$ is a contraction mapping is to show that the derivative is less than $1$ in the domain in question (it's a little bit tricky to prove that this implies that $f()$ is a contraction mapping, but not too hard, and it should hopefully be intuitively clear). Now, the derivative of $f$ is $f'(x)=\dfrac{2}{(2+x)^2}$, and this is clearly less than $1$ for any $x\geq 0$, so the conditions hold and the limit is well-defined.
As for why showing convergence matters in a problem like this, consider the expression $6-2\times(6-2\times(6-2\times(6-\ldots)))$; applying the same argument shows that the limit $\alpha$ satisfies $\alpha=6-2\alpha$, so $\alpha=2$ if the limit exists. But iterating the map $x\mapsto 6-2x$ blows up to infinity for any starting value of $x$, so the original expression doesn't have any proper meaning (over the real numbers).
A: 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.
$1 + x = \sqrt 2$
