My title may have come off as informal or nonspecific. But, in fact, my title could not be more specific.

Define a sequence with initial term: $$S_0=1+\frac{1}{1+\frac{1}{1+\frac1{\ddots}}}$$ It is well-known that $S_0=\frac{1+\sqrt{5}}{2}$. I do not write it as such because, in order to describe the next term, we must look at its continued fraction representation. Take every $1$ not a numerator and replace it with $\frac1{1+\frac1{\ddots}}$ to get: $$S_1=\frac1{1+\frac1{1+\frac1{\ddots}}} +\frac{1}{\frac1{1+\frac1{1+\frac1{\ddots}}}+\frac1{\frac1{1+\frac1{1+\frac1{\ddots}}}+\frac1\ddots}}$$ Repeat to get: $$S_2=\frac1{\frac1{1+\frac1{1+\frac1{\ddots}}}+\frac1\ddots}+\frac{1}{\frac1{\frac1{1+\frac1{1+\frac1{\ddots}}}+\frac1\ddots}+\frac1{\frac1{\frac1{1+\frac1{1+\frac1{\ddots}}}+\frac1\ddots}+\frac1{\ddots}}}$$ One more time, for good measure: $$S_3=\frac{1}{\frac1{\frac1{1+\frac1{1+\frac1{\ddots}}}+\frac1\ddots}+\frac1{\frac1{\frac1{1+\frac1{1+\frac1{\ddots}}}+\frac1\ddots}+\frac1{\ddots}}}+\frac1{\frac{1}{\frac1{\frac1{1+\frac1{1+\frac1{\ddots}}}+\frac1\ddots}+\frac1{\frac1{\frac1{1+\frac1{1+\frac1{\ddots}}}+\frac1\ddots}+\frac1{\ddots}}}+\frac1{\frac{1}{\frac1{\frac1{1+\frac1{1+\frac1{\ddots}}}+\frac1\ddots}+\frac1{\frac1{\frac1{1+\frac1{1+\frac1{\ddots}}}+\frac1\ddots}+\frac1{\ddots}}}+\frac1{\ddots}}}$$ We buzz like fat bees to the question:

$$\text{What is } \lim_{n \to \infty}S_n\text{?}$$

**My almost-solution:** Define a recursive sequence as follows:
$$\eta_0 = 1$$
$$\eta_k = \frac1{\eta_{k-1}+\frac1{\eta_{k-1}+\frac1\ddots}}$$
Notice:
$$\eta_k = \frac1{\eta_{k-1}+\eta_k} \implies \eta_k\eta_{k-1}+\eta_k^2-1=0 \implies \eta_k=\frac{-\eta_{k-1}+\sqrt{\eta_{k-1}^2+4}}2$$
We force the square root in the numerator to be positive for obvious reasons. Now, notice:
$$S_0 = \eta_0+\eta_1 = \varphi$$
$$S_1 = S_0-\eta_0+\eta_2=\eta_1+\eta_2$$
$$S_2 = S_1-\eta_1+\eta_3 = \eta_2 + \eta_3$$
$$S_3 = S_2-\eta_2+\eta_4 = \eta_3 + \eta_4$$
$$\vdots$$
$$S_n = S_{n-1} - \eta_{n-1} + \eta_{n+1} = \eta_n + \eta_{n+1}$$
By how I defined $\eta_i$, we may rewrite the RHS as:
$$S_n = \eta_n + \frac{-\eta_n+\sqrt{\eta_n^2+4}}2 \implies S_n = \frac{\eta_n+\sqrt{\eta_n^2+4}}2$$
This may also be obtained by the equivalence (which I noticed visually):
$$\eta_k = \frac1{S_{k-1}}$$
For $k \geq 1$. I omit the proof of this, although it is simple. With this in mind, we see:
$$S_k = \frac{\eta_k+\sqrt{\eta_k^2+4}}2 \implies 2S_k = \eta_k+\sqrt{\eta_k^2+4}$$
$$\implies 2S_k = \frac1{\eta_{k-1}+\eta_k} + \sqrt{\frac1{(\eta_{k-1}+\eta_k)^2 }+4}$$
$$\implies 2S_k = \frac1{\frac1{S_{k-2}} + \frac1{S_{k-1}}} + \sqrt{\frac1{(\frac1{S_{k-2}} + \frac1{S_{k-1}})^2 }+4}$$
Which, as $k \to \infty$, becomes the sum of an infinitely nested fraction and the square root of the sum of the reciprocal of a square of an infinitely nested fraction and $4$. In theory, this can be evaluated. But I draw my final breath and collapse, dead.

*More seriously,* what I have done is defined $S_n$ recursively, with initial values $S_0 = \varphi$ and $S_1 = \varphi + \frac1{\varphi+\frac1{\ddots}}$. But I don't know if this converges. An inverse symbolic calculator might find something.

**Edit (Solution):** it converges to $\sqrt{2}$.