Continued fraction involving Fibonacci sequence

Solution 1:

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 $\in\bbox[5px,border:2px solid blue]{(\frac{900}{1530};\frac{901}{1530})}$