I do not mean the continued fraction representation of zeta function; I mean the function which has the form:

$$f(s)=\cfrac{1}{1^s+\cfrac{1}{2^s+\cfrac{1}{3^s+\cfrac{1}{4^s+\cdots}}}}$$

For some values, we know the function exactly:

$$\begin{align*} f(0)&=\phi-1=0.61803398875\ldots\\ f(1)&=\frac{I_1(2)}{I_0(2)}=0.69777465796\ldots\\ f(-1)&=\frac{\pi}{2}-1=0.5707963268\ldots \end{align*}$$

And obviously, for big positive $s$, it approaches $1$.

$$f(s \rightarrow +\infty) \rightarrow 1$$

Using the convergence criterion for simple continued fractions, we can conclude that $f(s<-1)$ doesn't converge, which is confirmed by the behavior of its approximants.

Here are the first few approximations, based on truncating the continued fraction:

$$\begin{align*} f_1(s)&=1\\ f_2(s)&=\frac{1}{1+2^{-s}}\\ f_3(s)&=\frac{2^s+3^{-s}}{1+2^s+3^{-s}}\\ f_4(s)&=\frac{1+2^s(3^s+4^{-s})}{1+(1+2^s)(3^s+4^{-s})} \end{align*}$$

Defining (thanks to vrugtehagel for his helpful comment):

$$f(s)=\lim_{n \rightarrow \infty} f_n(s)$$

How would you analyze this function? Can it be turned into a series? Which of its properties can be found theoretically? Maybe it was already studied, then I would like some reference.

**Edit**

Using recurrence relations for continued fractions, I was able to get approximants for $f(s)$ of any order. The branching point near $s=-1$ is easy to see.