Question

Prove that the difference of the infinite continued fractions $\frac{1}{a+}\frac{1}{b+}\frac{1}{c+}...,\frac{1}{b+}\frac{1}{a+}\frac{1}{c+}.....,$ is equal to $\frac{a-b}{1+ab}$.

Answer 1

Let $x=\frac{1}{a+}\frac{1}{b+}\frac{1}{c+}.....$

$\therefore x=\frac{1}{a+}\frac{1}{b+}\frac{1}{c+x}=\frac{b(c+x)+1}{(ab+1)(c+x)+a};$

On reduction we obtain;-

$(1+ab)x^2+(abc+a+c-b)x-(1+bc)=0$

Similarly, $y=\frac{1}{b+}\frac{1}{a+}\frac{1}{c+y}$

Hence, $(1+ab)y^2+(abc+b+c-a)y-(1+ac)=0$

Subtracting and rearranging, we have

$(1+ab)(x^2-y^2)+(ab+1)c(x-y)+(a-b)(x+y)+c(a-b)=0$

i.e.,

$(1+ab)(x-y)(x+y+c)+(a-b)(x+y+c)=0$

Now $x+y+c$ is positive quantity,

Hence, $(1+ab)(x-y)+(a-b)=0;$

$\therefore x-y=\frac{b-a}{1+ab}$

Hence, the difference is;-

$=\frac{a-b}{1+ab};$