a certain simple continued fraction Given the golden ratio: $$\phi=\frac{1+\sqrt{5}}{2}$$ and the following simple continued fraction:
$$G(q,k)=\cfrac{1}{1-q+\cfrac{1}{1-{q^3}^k+\cfrac{1}{1-{q^5}^k+\cfrac{1}{1-{q^7}^k+\ddots}}}}$$
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 $$\frac{1}{\phi-q}$$ 
 A: 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.
A: 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,
$$G(q,k)\approx\cfrac{1}{1-q+\cfrac{1}{1-{q^3}^k+\cfrac{1}{1-{q^5}^k+\cfrac{1}{1-{q^7}^k}}}}$$
is, 
$$G(q,k)\approx \frac{3}{5}+\frac{3^2}{5^2}q+\frac{3^3}{5^3}q^2+\ddots$$
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.
