Question

Prove that $\left(a+\frac{1}{b+}\frac{1}{a+}\frac{1}{b+}\frac{1}{a+}....\right)\left(\frac{1}{b+}\frac{1}{a+}\frac{1}{b+}\frac{1}{a+}....\right)=\frac{a}{b}$.

Answer 1

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

$=a+\frac{1}{b+\frac{1}{a+x-a}}=a+\frac{x}{bx+1}$

Thus $b{x}^2+x=abx+x+a$

$\Rightarrow b{x}^2-abx-a=0$

$\Rightarrow x=\frac{ab+\sqrt{a^2{b}^2+4ab}}{2b}$

$\Rightarrow y=\frac{1}{b+\frac{1}{a+y}}=\frac{a+y}{ab+by+1}$

$\Rightarrow aby+b{y}^2+y=a+y$

$\Rightarrow b{y}^2+aby-a=0$

$\Rightarrow y=\frac{-ab+\sqrt{a^2{b}^2+4ab}}{2b}$

$\Rightarrow xy=\frac{a^2{b}^2+4ab-a^2{b}^2}{4b^2}$

$=\frac{a}{b}$

Hence, proved.