Question

Prove that the $n^{th}$ convergent to the continued fraction
$\frac{r}{r+1-}\frac{r}{r+1-}\frac{r}{r+1-}\cdots$ is $\frac{r^{n+1}-r}{r^{n+1}-1}$.

Answer 1

We have $u_n=(r+1)u_{n-1}-r{u}_n;$

$u_n-(r+1)u_{n-1}+r{u}_n=0$

$u_1+u_{2}x+.....$

Thus the series $1-(r+1)x+r{x}^2$ is a recurring series, whose scale of relation is $\frac{u_1+\{u_1+(r+1)u_1\}x}{1-x-x^2}$ and whose generating function is,

Now, $p_1=r,q_1=r+1,p_2=r(r+1),q_2=r^2+r+1$

$\therefore p_1+p_{2}x+p_3{x}^2+.....=\frac{r}{1-(r+1)x+r{x}^2}$

$=\frac{r}{(r-1)}(\frac{r}{1-rx}-\frac{1}{1-x})$

$p_n=\frac{r}{r-1}(r^n-1)$

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

$q_n=\frac{1}{r-1}(r^{n+1}-1)$

$\frac{p_n}{q_n}=\frac{r(r^n-1)}{r^{n+1}-1}$