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!!

  • 204,278
  • 154
  • 264
  • 488
Carlos Toscano-Ochoa
  • 2,505
  • 1
  • 12
  • 26
  • 2
    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. – p4sch Sep 03 '18 at 19:20
  • Since it's an infinite [regular continued fraction](http://mathworld.wolfram.com/RegularContinuedFraction.html) it converges to an irrational number. – saulspatz Sep 03 '18 at 19:20
  • 1
    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 03 '18 at 19:21
  • 1
    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 03 '18 at 19:26
  • It is the sequence [A135829](https://oeis.org/A135829)(n)/[A026822](https://oeis.org/A026822)(n), which at least allow you to express it as a recurrence, but nothing nice seems to come from it. – Sil Sep 04 '18 at 21:15
  • Definitely an interesting question, but I'm not sure that it's really an algebraic number theory question. – Robert Soupe Sep 05 '18 at 00:44
  • If the plus signs were replaced by alternating $\mp$, the CF could be rewritten as $\frac{1}{\sqrt{5}}{\large\mathrm{K}}_{n=1}^\infty\frac{a q^n}{1 - q^n +{}}$ with $a=-5\Phi$, $q=-\Phi^{-2} = \Phi - 2$. According to Ramanujan's 2nd notebook, this would then be equal to $\frac{1}{\sqrt{5}}\left(\frac{F(a,q)}{F(aq,q)}-1\right)$ where $F(a,q) = \sum_{n=0}^\infty\frac{a^n q^{n^2}}{(q;q)_n^2}$. – ccorn Sep 28 '18 at 00:40

1 Answers1


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