Question

If $u_1=\frac{a}{b},u_2=\frac{b}{a+b},u_3=\frac{a+b}{a+2b},\cdots$, each successive fraction being formed by taking the denominator and the sum of the numerator and denominator of the preceding fraction for its numerator and denominator respectively, show that $u_{\infty }=\frac{\sqrt{5}-1}{2}$.

Answer 1

Here, $u_1=\frac{p_1}{q_1},u_2=\frac{p_2}{q_2},u_3=\frac{p_3}{q_3},.....$

Where, $p_n=q_{n-1}$ and $q_n=q_{n-1}+p_{n-1}$

$q_n=q_{n-1}+q_{n-2}$

Hence,, $q_1+q_{2}x+q_3{x}^2+....$ is a recurring series in which the scale of relation is $1-x-x^2$

$q_1=b,q_2=a+b$

$q_1+q_{2}x+q_3{x}^2+....=\frac{q_1+(q_2-q_1)x}{1-x-x^2}$

$=\frac{b+ax}{1-x-x^2}$

Let $\frac{b+ax}{1-x-x^2}=\frac{A}{1-\alpha x}+\frac{B}{1-\beta x};$

Then $q_n=A{\alpha }^{n-1}+B{\beta }^{n-1};p_n=q_{n-1}=A{\alpha }^{n-2}+B{\beta }^{n-2}$

$\therefore \frac{p_n}{q_n}=\frac{A{\alpha }^{n-2}+B{\beta }^{n-2}}{A{\alpha }^{n-1}+B{\beta }^{n-1}}$

Now, $\alpha$ and $\beta$ are to be found from the equations $\alpha +\beta =1,\alpha \beta =-1$,

Let $\alpha$ be the greater of the quantities; then $\alpha =\frac{1+\sqrt{5}}{2};\beta =\frac{1-\sqrt{5}}{2}$

So that $\alpha >1,\beta <1;$ hence the limit when $n$ is infinite of ${\alpha }^n$ is infinity and ${\beta }^n$ is $0$.

$\frac{p_n}{q_n}=\frac{A{\alpha }^{n-2}}{A{\alpha }^{n-1}}=\frac{1}{\alpha }=\frac{2}{1+\sqrt{5}}$ $=\frac{\sqrt{5}-1}{2}$