Given the golden ratio: $$\phi=\frac{1+\sqrt{5}}{2}$$ and the following simple continued fraction:


For $|q|\lt1$, prove/disprove that the continued fraction converges to the limit $$\frac{\phi}{1-{\phi}q}$$ as $k\to\infty$.

As k becomes finitely large,the cfrac seems to approach the limit.

(Edit):it was a mistake on my part, I misinterpreted the reciprocal of phi as phi.The limit Is indeed $$\frac{1}{\phi-q}$$

  • $\begingroup$ Since we are given that $k\in\mathbb N$, the part "iff $1\le k<\infty$ seems redundant, doesn't it? $\endgroup$ – Hagen von Eitzen Aug 31 '15 at 5:45
  • $\begingroup$ @Hagon Von Eltzen I guess it is clear now, $\endgroup$ – Nicco Aug 31 '15 at 5:53
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    $\begingroup$ I think one does need to say "iff $k \in \mathbb{N}$" somehow, provided it is intended that the equality be false when $k$ is not a natural number. $\endgroup$ – coffeemath Aug 31 '15 at 6:18
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    $\begingroup$ @Nicco: I think it will help your post, and your discussion with coffeemath, if you give a concrete example of a $q$ such that $G(q,k) = \frac{\phi}{1-\phi q}$ as $k\to \infty$. $\endgroup$ – Tito Piezas III Aug 31 '15 at 12:10
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    $\begingroup$ @Nicco: as Lucian says, if $|q|\lt1$, then $q^{a^k}\to0$, so the continued fraction is tending to $$\cfrac1{1-q+\cfrac1{\color{#C00000}{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}}}}$$ where the red part is the continued fraction for $\phi$. Then $\frac1\phi=\phi-1$. $\endgroup$ – robjohn Aug 31 '15 at 19:34

Take $k=1$ and $q=1/2.$ Then the first few convergents are (starting with the trivial zeroth which here is $0$) $$C(0)=0,\\ C(1)=2,\\ C(2)=14/23=0.60869..,\\ C(3)=946/969=0.97626..,\\ C(5)=177486/217271=0.81688.$$ Since the terms $b(k)=1-(1/2)^{2k-1}$ go so rapidly to $1$ it would seem odd (to me) if the convergents were not alternately below/above the value of the fraction. [Note I haven't poved this.] Given that, it seems we have convergence to something certainly between $0.82$ and $0.97.$

However your formula predicts $\phi/(1-\phi \cdot (1/2)),$ about $8.47213..$ so to me it seems your formula is off somewhere. I tried varying the formula but couldn't get a match. It also seems unlikely (to me) that there be no occurrence of the parameter $k$ in such a formula.

About taking the limit as $k \to \infty.$ Suppose we again use $q=1/2$ so that the predicted formula gives about $8.47$ as before. The term used for the continued fraction is $b(j,k)$ where $$b(j,k)=1-(1/2)^{(2j-1)^k},$$ and where $k$ is fixed and the index $j$ is for the continued faction $[0;b(1,k),b(2,k),\cdots]$ in usual continued fraction notation. Note that $b(1,k)=1-(1/2)^1=1/2$ independent of $k$, so with the zeroth and first convergents being $0,2$ we expect the fraction to converge to something in the interval $(0,2),$ provided convergents alternate above and below. The second convergent is the smallest which uses the parameter $k,$ and we have $b(2,k)=1-(1/2)^{3*k},$ and if simplified we get the second convergent $$\frac{2(2^{3^k}-1)}{3 \cdot 2^{3^k}-1}.$$ This can be seen to approach $2/3$ (rapidly) as $k \to \infty,$ or by algebraic rearrangement it is $$\frac{2}{3}(1-\frac{2/3}{2^{3^k}-1/3})$$ making clear the approach to $2/3$.$ [My polydigit calculator didn't do well getting approximations directly for this]

Anyway it seems the limit as $k \to \infty$ of the continued fraction is somewhere between $2/3$ and $2$ (relying on the alternately above/below behavior, which again I have not proved for this). If this is right it is nowhere near $8.$

Just for clarity, I followed up in chat with the OP and did note the $1/(\phi-q)$ value. A fun cfrac, IMO.

  • $\begingroup$ @ coffeemath As you increase your $k$,say to $k=5$,you'll get better approximation. $\endgroup$ – Nicco Aug 31 '15 at 9:39
  • $\begingroup$ @Nicco If you mean to take the limit of your continued fraction as $k \to \infty,$ then that should be stated in the posted question. As it seems, the question claims a result independent of $k.$ $\endgroup$ – coffeemath Aug 31 '15 at 10:13
  • $\begingroup$ @Nicco See the last added part of my answer, I think for k to infinity the cfrac is still somewhere between 2/3 and 2. $\endgroup$ – coffeemath Aug 31 '15 at 11:29
  • $\begingroup$ given the formula,for example in the 3th convergent,we replace phi by (2/3) ,in the 4th by (3/5) and in the 5th by (5/8) ,which reveals the pattern of fibonacci numbers.Now going to infinity we'll have phi since the ratio of fibonacci numbers approaches phi $\endgroup$ – Nicco Aug 31 '15 at 11:49
  • $\begingroup$ @Nicco -- I don't follow the last comment. The cfrac has no position in which to replace $\phi$ by various things, only your final formula has $\phi$ in it. In the cfrrac itself there are only the terms $1-q,1-q^{3^k},$ and so on, As $k \to \infty$ all of these except for the first one, with no $k$ in it, will approach $1$ making the whole thing go to $[0,(1-q),1,1,...]$ or $1/(1-q+\phi-1)=1/(\phi-q),$ which for $q=1/2$ is $0.89442...$. $\endgroup$ – coffeemath Aug 31 '15 at 13:38

following Tito Piezas III comment I'll provide some concrete examples. First, let's take the second convergent of the cfrac

$$G(q,k)\approx\cfrac{1}{1-q+\cfrac{1}{1-{q^3}^k}}$$ and expand it into a power series as $k\to\infty$

$$G(q,k)\approx \frac{1}{2}+\frac{1}{2^2}q+\frac{1}{2^3}q^2+\dots$$

Which converges to $\frac{2}{3}$ for $q=\frac{1}{2}$ as shown by Coffeemath and also by the geometric series formula $$\frac{a}{1-aq}$$ with $a=\frac{1}{2}$. Taking the third convergent,we have

$$G(q,k)\approx\cfrac{1}{1-q+\cfrac{1}{1-{q^3}^k+\cfrac{1}{1-{q^5}^k}}}$$ and its power series for $k\to\infty$

$$G(q,k)\approx \frac{2}{3}+\frac{2^2}{3^2}q+\frac{2^3}{3^3}q^2+\ddots$$

Which converges to $1$. The fourth convergent,



$$G(q,k)\approx \frac{3}{5}+\frac{3^2}{5^2}q+\frac{3^3}{5^3}q^2+\ddots$$

Doing this ad infinitum ,we observe that the ratio of the geometric series is always a ratio of two consecutive fibonacci numbers ,for each convergent.

Taking the limit of the cfrac to infinity ,we are led to conjecture the given limit.

  • $\begingroup$ Kindly be careful with your formatting. A post looks nicer if the spacing is not over-done. :) $\endgroup$ – Tito Piezas III Aug 31 '15 at 14:36
  • $\begingroup$ @ Tito Piezas Thanks a lot $\endgroup$ – Nicco Aug 31 '15 at 14:52
  • $\begingroup$ Nicco -- It seems now that if you just use the sums of your geometric series for the convergents, (where you have already let $k \to \infty$) then you get $C(n)=F_n/(F_{n+1}-F_nq)$ for the $n$th convergent. If the top and bottom here are divided by $F_n$ it is seen to approach $1/(\phi-q)$ as I noted in a comment to my answer. This is only off by missing a $\phi$ from your formula, since it can also be written as $\phi / (1+\phi-\phi q).$ $\endgroup$ – coffeemath Aug 31 '15 at 21:10
  • $\begingroup$ @ coffeemath thanks for your efforts,the limit you suggested is indeed true $\endgroup$ – Nicco Aug 31 '15 at 21:31

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