<s>[INST] Inequality related to the continued fraction expansion of $\sqrt 3$ I am working on a problem related to the continued fraction expansion of $\sqrt 3$. If $p_k$ and $q_k$ denote the numerator and denominator, respectively, of the $k$th convergent, I should show that
$$\left|\sqrt{3}-\frac{p_{2n+1}}{q_{2n+1}}\right| \lt \frac{1}{2\sqrt{3}q_{2n+1}^2}\;.$$
I have determined that the continued fraction expansion of $\sqrt 3$ is $[1;\overline{1,2}]$ and I am able to show the equality apart from the factor $2\sqrt3$ in the denominator.
Any suggestions?
Thanks in advance!
 [/INST] From the period length of $2$, you obtain that for all $n$, you have
$$p_{2n+1}^2 - 3q_{2n+1}^2 = 1,$$
and from that you can deduce
$$\begin{align}
\frac{p_{2n+1}}{q_{2n+1}} - \sqrt{3} &= \frac{p_{2n+1} - \sqrt{3}q_{2n+1}}{q_{2n+1}}\\
&= \frac{(p_{2n+1} - \sqrt{3}q_{2n+1})(p_{2n+1} + \sqrt{3}q_{2n+1})}{q_{2n+1}(p_{2n+1} + \sqrt{3}q_{2n+1})}\\
&= \frac{1}{\left(\frac{p_{2n+1}}{q_{2n+1}}+\sqrt{3}\right)q_{2n+1}^2}\\
&< \frac{1}{2\sqrt{3}q_{2n+1}^2}
\end{align}$$
since $\frac{p_{2n+1}}{q_{2n+1}}>\sqrt{3}$ (which you can either deduce from the general fact that odd convergents are larger than [or equal to] the value of the continued fraction, or from the fact mentioned above).
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