Calculate the exact value of an infinite sequence $$1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{...}}}$$ I think I need to convert that sequence to sequence in terms of n and then evaluate $$\lim \limits_{n \to \infty}$$ but I'm really stuck into converting it.
 A: By noting that $…=r$, then I would start from the end
$$\frac{1}{…}=\frac{1}{r}$$
$$1+\frac{1}{…}=1+\frac{1}{r}=\frac{r+1}{r}$$
Skipping one step,
$$1+\frac{1}{1+\frac{1}{…}}=1+\frac{1}{1+\frac{1}{r}}=\frac{2r+1}{r+1}$$
Next steps render $$s_3=\frac{3r+2}{2r+1}$$
$$s_4=\frac{5r+3}{3r+2}$$
$$s_n=\frac{f_n \cdot r+f_{n-1}}{f_{n-1}\cdot r +f_{n-2}}$$, where $f_n$ are the terms of the Fibonacci sequence.
A: Like @mowzorn said, it is a good idea to let $$x=1+\frac{1}{1+\frac{1}{1+\ldots}}$$ so that $x=1+\frac{1}{x}$.
You will see that there are two values of $x$ for which this works - $\frac{1+\sqrt5}{2}$ and $\frac{1-\sqrt5}{2}$. It seems strange that this expression could have two different values. You can argue that only $\frac{1+\sqrt5}{2}$ is a valid solution but you could also argue that $\frac{1-\sqrt5}{2}$ is a solution to your problem. 3Blue1Brown made a video on this, using differentiation to show the two perspectives - the link is here: The other way to visualize derivatives | Chapter 12, Essence of calculus.
