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 $\Phi$?

Many thanks in advance!!

  • 2
    $\begingroup$ All 'regular' continued fractions are convergent and if they are also infinite, then the limit is irrational. This can be found in any (good) book on elementary number theory. $\endgroup$
    – p4sch
    Sep 3, 2018 at 19:20
  • $\begingroup$ Since it's an infinite regular continued fraction it converges to an irrational number. $\endgroup$
    – saulspatz
    Sep 3, 2018 at 19:20
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 3, 2018 at 19:21
  • 1
    $\begingroup$ I tried with the Inverse Symbolic Calculator, who said "Wow, really found nothing." So it probably can't be easily expressed in terms of $\Phi$. $\endgroup$
    – Kusma
    Sep 3, 2018 at 19:26
  • $\begingroup$ It is the sequence A135829(n)/A026822(n), which at least allow you to express it as a recurrence, but nothing nice seems to come from it. $\endgroup$
    – Sil
    Sep 4, 2018 at 21:15

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


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