The monster continued fraction

Solution 1:

It's not too hard to show** the sequence $\eta_k\to\frac{1}{\sqrt{2}}$ as $k\to\infty$, implying your desired $k\to\infty$ limit of $S_k=\eta_k^{-1}$ is $\sqrt{2}$. There is an error in your obtaining a recursion relation for the sequence $S_k$: it should read$$\color{limegreen}{2}S_k=\frac{1}{\frac{1}{S_{k-2}}+\frac{1}{S_{k-1}}}+\sqrt{\frac{1}{(\frac{1}{S_{k-2}}+\frac{1}{S_{k-1}})^2}+4}.$$One could also obtain the behaviour of $S_n$ from $\eta_n+\eta_{n+1}$.

** The convergence is linear. Writing $\eta_k=\frac{1}{\sqrt{2}}+\epsilon_k$ gives $\eta_k^2+4=\frac92+\epsilon_k\sqrt{2}+\epsilon_k^2$ and $\sqrt{\eta_k^2+4}=\frac{3}{\sqrt{2}}+\frac13\epsilon_k+O(\epsilon_k^2)$, whence $\eta_{k+1}=\frac{1}{\sqrt{2}}-\frac13\epsilon_k+O(\epsilon_k^2)$ and $S_k=\eta_k+\eta_{k+1}=\sqrt{2}+\frac23\epsilon_k+O(\epsilon_k^2)$.

Solution 2:

It took me quite a while to wrap my head around what you actually meant here (especially since $1$ is essentially used as a variable). Here is my attempt at making this rigorous, as I found the other answers here lacking. This isn't really a question about continued fractions, so let us start by doing away with them.

I will need the Banach fixed-point theorem, which I will state in a very special case:

Banach fixed-point theorem

Suppose that $I \subset \mathbb{R}$ is a closed interval (not necessarily bounded), and that $f \colon I \to I$ is a contraction. That is, there is a constant $\alpha \in (0,1)$ such that $$|f(x)-f(y)|\leq \alpha |x-y|$$ for all $x,y \in I$. Then there is a unique (fixed) point $a \in I$ such that $f(a) = a$, and moreover $f^n(x) \to a$ for every $x \in I$.

(Here $f^n$ means $n$-fold function composition. For instance, $f^3(x) = (f \circ f \circ f)(x) = f(f(f(x)))$.)

In fact, I will need a slight generalization of this, namely

Banach fixed-point theorem (generalized)

Suppose that $I \subset \mathbb{R}$ is a closed interval, that $f \colon I \to I$ is some function on $I$, and that $f^k$ is a contraction for some $k \geq 1$. Then $f$ has a unique fixed point $a$, and moreover $f^n(x) \to a$ for every $x \in I$.

Neither of these are difficult to prove, but let's not reinvent the wheel ;)

First off, what does the expression $$[\beta;\beta,\beta,\ldots] = \beta + \frac{1}{\beta+\frac{1}{\beta +\frac{1}{\ddots}}}$$ even mean? It means the limit of the sequence with terms (that is, stopping the continued fraction after $n$ levels) $$a_n(\beta) = [\beta;\underbrace{\beta,\beta,\ldots,\beta}_\text{$n$ times}],$$ assuming this limit exists. Thinking of $\beta > 0$ as fixed, we define $f_\beta \colon [\beta,\infty) \to [\beta,\infty)$ through $$f_\beta(x) := \beta + \frac{1}{x},$$ and observe that, crucially, $$a_n = \beta + \frac{1}{a_{n-1}} = f_\beta(a_{n-1}) =f_\beta^2(a_{n-2}) = \cdots = f_\beta^{n}(a_0) = f_\beta^{n}(\beta).$$

While $f_\beta$ itself may not be a contraction (depending on the size of $\beta$), you can easily compute that $$(f_\beta^2)'(x) = \frac{1}{(1+\beta x)^2}.$$ Thus $f_\beta^2$ is a contraction with contraction constant $(1+\beta^2)^{-2} < 1$ by the mean value theorem. As a consequence of the second fixed-point theorem above, $f_\beta$ has a unique fixed point, let's say $a(\beta)$. And moreover, $a_n(\beta) = f_\beta^n(\beta) \to a(\beta)$. As has been mentioned in several other comments, a simple quadratic equation shows that, in fact $$a(\beta) = \frac{\sqrt{\beta^2 + 4}+\beta}{2}.$$

The second continued fraction we need comes for free from the above, namely $$g(\beta) := \frac{1}{\beta + \frac{1}{\beta + \frac{1}{\ddots}}} = a(\beta) - \beta = \frac{\sqrt{\beta^2 + 4}-\beta}{2}.$$

The functions $a,g$ above are well defined $(0,\infty) \to (0,\infty)$, and your $S_n(\beta)$ (with $\beta = 1$ in your case) is in fact precisely $$S_n(\beta) = a(g^n(\beta)).$$ Now, we want to (yet again) apply Banach's fixed point theorem, but there is a slight technical issue of $(0,\infty)$ not being closed. We solve this by noting that for all $\beta \in [\varphi^{-1},1]$ we have $$g(\beta) = \frac{\beta^2+4-\beta^2}{2(\sqrt{\beta^2+4}+\beta} = \frac{2}{\sqrt{\beta^2+4}+\beta} \leq 1$$ and $$g(\beta) \geq \frac{2}{\sqrt{1+4}+1} = \varphi^{-1},$$ so we may consider $g$ to be a function $[\varphi^{-1},1] \to [\varphi^{-1},1]$ instead. Since $$g'(\beta) = \frac{1}{2}\left(\frac{\beta}{\sqrt{\beta^2+4}}-1\right)$$ satisfies $-1/2 < g'(\beta) < 0 $, we get that $g$ is a contraction (with constant $1/2$). It therefore has a unique fixed point $\gamma \in [\varphi^{-1},1]$, which you can easily compute to be $\gamma = 1/\sqrt{2}$, and more importantly $g^n(\beta) \to \gamma$ for all $\beta \in [\varphi^{-1},1]$.

Finally, using the continuity of $a$, we conclude that $$S_n(\beta) =a(g^n(\beta)) \to a\left(\frac{1}{\sqrt{2}}\right) = \sqrt{2}$$ for every $\beta \in [\varphi^{-1},1]$, and in particular for $\beta \equiv 1$.

(In fact, using a more quantitative version of Banach's fixed point theorem, it is not difficult to show more explicitly that $|S_n(1)-\sqrt{2}| \leq 2^{1-n}(1-\varphi^{-1})$, but this post is already significantly longer than I expected...)

Solution 3:

Assume that $(S_n)_{n\geq 0}$ converges to a finite limit $x\geq 0$. Taking limits in your final equality yields (this works also if $x=0$) $x=\frac{x}{2}+\sqrt{\dfrac{x^2}{4}+4}$. Hence $\frac{x}{2}=\sqrt{\dfrac{x^2}{4}+4}$, and squaring gives you $0=4$.

Thus, $(S_n)_n$ is divergent.

Edit. From my comments, and as J.G.'s answer confirms, there was some mistake in the recursion ,and $S_n$ converges to $\sqrt{2}$.

Solution 4:

Hmm. I get $$ x={1\over x+{1\over x + \cdots}}={1\over x+ x}={1\over 2x} \qquad \text{if this converges}$$ Then $$2x^2=1 \\ x=\sqrt{1/2} $$ Isn't it so simple?