# Continued fraction involving Fibonacci sequence

What is the limit of the continued fraction:

$$\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{5+\cfrac{1}{8+\cdots}}}}}}\$$

that involves the Fibonacci sequence terms as denominators? I've been looking for this specific continued fraction on the Internet, but I haven't been able to find anything about it. Does it converge? Is it algebraic or transcendental? Is it somehow a relevant constant? Can it be expressed in terms of $\Phi$?

• If you have all $1$'s as the addends, then the fraction becomes one divided by the golden ratio and its approximants are ratios of each Fibonacci number to the next one. – Oscar Lanzi Sep 3 '18 at 19:21
• I tried with the Inverse Symbolic Calculator, who said "Wow, really found nothing." So it probably can't be easily expressed in terms of $\Phi$. – Kusma Sep 3 '18 at 19:26
I can answer the converge aspect: this fraction is tricky - we have to be careful: $$1=\cfrac{1}{1}>\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{5+\cfrac{1}{8+\cdots}}}}}}>\frac{1}{1+something_{Big}}\approx0$$ $$\Phi>\frac35=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2}}}>\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{5+\cfrac{1}{8+\cdots}}}}}}>\cfrac{1}{1+\cfrac{1}{1}}=\frac12$$ $$\frac{53}{90}=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{5}}}}}>\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{5+\cfrac{1}{8+\cdots}}}}}}>\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3}}}}=\frac{10}{17}$$
I think we can stop here: the result $$\in(\frac{10}{17};\frac{53}{90})<\Phi$$ . Is same to say $$\in\bbox[5px,border:2px solid blue]{(\frac{900}{1530};\frac{901}{1530})}$$