TITLE: $\sqrt{11}$ as rational number within $10^{-4}$?
QUESTION [5 upvotes]: Use continued fractions to find a rational number which approximates $\sqrt{11}$ to
within $10^{−4}$. 
I know how to solve for continued fractions like this:
$$\sqrt{11}=3+x$$
$$11=9+6x+x^2$$
$$11=9+(6+x)x$$
$$2=x(6+x)$$
$$x=\frac{2}{6+x}=\frac{1}{3+\frac{x}2}=\frac{1}{3+\frac{1}{6+x}}$$
therefore, $$\sqrt{11}=[3;\overline{3,6}]$$
How do I find out where to terminate the fraction so as to obtain close value with error less than  the mentioned limit

REPLY [3 votes]: Theorem: Let $p_k/q_k$ is the $k$-th convergent of a positive real $x$
  then 
$$ \Bigg|x - \frac{p_k}{q_k}\Bigg| \le \frac{1}{q_k q_{k+1}} <
\frac{1}{q_k^2} $$

So to approximate within $10^{-4}$, $q_k > 100$ sufficient not always necessary because in general you can get the same accuracy with a smaller denominator if $q_k q_{k+1} > 10^4$.