# Two equivalent simple continued fractions $A\overset{\color{red}?}=B$

Fix a sequence $a_n$ of complex numbers for natural number $n$,such that $|a_n|\gt1$

and then define the following two simple continued fractions

$A= \cfrac{1}{a_{0}-\cfrac{1}{a_{1}+\cfrac{1}{a_{2}-\cfrac{1}{a_{3}+\cfrac{1}{a_{4}-\cfrac{1}{a_{5}+\cfrac{1}{a_{6}-\cfrac{1}{a_{7}+\dots}}}}}}}}\tag1$

which converges by Śleszyński-Pringsheim Theorem

and

$B= \cfrac{1}{a_{0}-1+\cfrac{1}{1+\cfrac{1}{a_{1}-1+\cfrac{1}{a_{2}-1+\cfrac{1}{1+\cfrac{1}{a_{3}-1+\cfrac{1}{a_{4}-1+\cfrac{1}{1+\cfrac{1}{a_5-1+\cfrac{1}{a_{6}-1+\cfrac{1}{1+\dots}}}}}}}}}}}\tag2$

How do we prove that $A=B$ ?

Special case

(i) Define $R(q)=q^{1/5}\frac{(q;q^5)_{\infty}(q^4;q^5)_{\infty}}{(q^2;q^5)_{\infty}(q^3;q^5)_{\infty}}$ the Rogers-Ramanujan continued fraction where $|q|\lt1$ ,then

$R(q)=\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cfrac{q^4}{1+\cfrac{q^{5}}{1+\dots}}}}}}$

and

$R(q)=\cfrac{q^{1/5}}{1+\cfrac{q}{1-q+\cfrac{q}{1-\cfrac{q}{1+q+\cfrac{q^3}{1+q^2-\cfrac{q^2}{1+\cfrac{q^2}{1-q^2+\cfrac{q^5}{1-q^3+\cfrac{q^3}{1-\cfrac{q^3}{1+q^3+\cfrac{q^7}{1+q^4-\cfrac{q^4}{1+\dots}}}}}}}}}}}}$

(ii) Let $a_n=2$,then

$A= \cfrac{1}{2-\cfrac{1}{2+\cfrac{1}{2-\cfrac{1}{2+\cfrac{1}{2-\cfrac{1}{2+\dots}}}}}}=\frac{1-\sqrt{5}}{2}$

and

$B=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\dots}}}}}}=\frac{1-\sqrt{5}}{2}$

Remark: this general identity leads to another continued fraction for the Generalized Rogers-Ramanujan continued fraction

Edited 17 Oct 2017:I've just noticed that this oeis article also discusses the same continued fraction for the special case of the Generalized Rogers-Ramanujan continued fraction $R(-1,q)$

$$1-\frac1{1+\frac 1 {x-1}} = 1-\frac {x-1}{x-1+1} = 1-\frac{x-1}x = \frac {x-x+1}x = \frac 1x$$
And in particular, $$\frac 1 {a-1 + \frac 1 {1 + \frac 1 {b-1 + x}}} = \frac 1 {a-\frac 1{b+x}}$$