Question

Show that the $n^{th}$ convergent to $1-\frac{1}{4-}\frac{1}{4-}\cdots$ is equal to the ${(2n-1)}^{th}$ convergent to $\frac{1}{1+}\frac{1}{2+}\frac{1}{1+}\frac{1}{2+}\cdots$.

Answer 1

The convergents to $\frac{1}{1+}\frac{1}{2+}\frac{1}{1+}\frac{1}{2+}.....$ are

$\frac{1}{1},\frac{2}{3},\frac{3}{4},\frac{8}{11},\frac{11}{15},\frac{30}{41},\frac{41}{56},$

and the convergents to $1-\frac{1}{4-}\frac{1}{4-},.....$ are

$\frac{1}{1},\frac{3}{4},\frac{11}{15},\frac{41}{56},\frac{153}{209},....$

Let $\frac{p_1}{q_1},\frac{p_2}{q_2},\frac{p_3}{q_3},,....;\frac{r_1}{s_1},\frac{r_2}{s_2},\frac{r_3}{s_3},$ denote the two sets of convergents;

then $p_1=r_1,p_3=r_2,p_5=r_3,p_7=r_4.....$ and similarly for q and s;

$p_{2n-1}=p_{2n-2}+p_{2n-3};$

$p_{2n-2}=p_{2n-3}+p_{2n-4};$

$p_{2n-3}=p_{2n-4}+p_{2n-5};$

Hence, $p_{2n-1}-4p_{2n-3}+p_{2n-5}=0;$

$r_n-4r_{n-1}+r_{n-2}=0$

thus, $p_9=4p_7-p_5=4r_4-r_3=r_5$

$p_{11}=4p_9-p_7=4r_5-r_4=r_6$

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Hence generally,

$p_{2n-1}=r_n$

Similarly, $q_{2n-1}=s_n$

$\frac{r_n}{s_n}=\frac{p_{2n-1}}{q_{2n-1}}$