Question

Find the $n^{th}$ convergent to $\frac{6}{5-}\frac{6}{5-}\frac{6}{5-}\cdots$.

Answer 1

We have $p_n=5p_{n-1}-6p_{n-2}$; hence the numerators form a recurring series any three consecutive terms of which are connected by the relation
$p_n-5p_{n-1}+6p_{n-2}$.
Let $S=p_1+p_{2}x+p_3{x}^2+\cdots +p_n{x}^{n-1}+\cdots$;
then, we have $S=\frac{p_1+(p_2-5p_1)x}{1-5x+6x^2}$.
But the first two convergents are $\frac{6}{5}$, $\frac{30}{19}$;
$\therefore S=\frac{6}{1-5x+6x^2}=\frac{18}{1-3x}-\frac{12}{1-2x}$;
whence $p_n=18\cdot 3^{n-1}-12\cdot 2^{n-1}=6(3^n-2^n)$.
Similarly if $S^{{}^{\prime }}=q_1+q_{2}x+q_3{x}^2+\cdots +q_n{x}^{n-1}+\cdots$,
we find $S^{{}^{\prime }}=\frac{5-6x}{1-5x+6x^2}=\frac{9}{1-3x}-\frac{4}{1-2x}$;
whence $q_n=9\cdot 3^{n-1}-4\cdot 2^{n-1}=3^{n+1}-2^{n+1}$.
$\therefore \frac{p_n}{q_n}=\frac{6(3^n-2^n)}{3^{n+1}-2^{n+1}}$.