How to calculate this limit related to continued fractions?

63 Views Asked by At

Given that $\lim_{n\to\infty}p_n/q_n=\sqrt{2}$, how to calculate $\lim_{n\to\infty}q_n|p_n-\sqrt{2}q_n|$?

Here, $p_n$ and $q_n$ are defined by the continued fraction of $\sqrt{2}$, and $p_n=a_np_{n-1}+p_{n-2}$ and $q_n=a_nq_{n-1}+q_{n-2}$, where $a_n$ is the $n$-th number in the continued fraction representation $\sqrt{2}=[1,2,2,\cdots]$.

I did a little bit of calculation for the first few terms and found out that the result should be $\sqrt{2}/4$, but I cannot figure out how to prove this result. It seems that ordinary single variable or multivariable approaches do not work in this case. Thanks!

1

There are 1 best solutions below

0
On

Here it's easy to solve the recurrences in closed form: \begin{align} p_n&=\frac{(1+\sqrt2)^{n+1}+(1-\sqrt2)^{n+1}}{2} \\q_n&=\frac{(1+\sqrt2)^{n+1}-(1-\sqrt2)^{n+1}}{2\sqrt2} \end{align} (which makes $\lim\limits_{n\to\infty}q_n\big|p_n-q_n\sqrt2\big|=\frac1{2\sqrt2}$ obvious).