Convergents of an infinite continued fraction oscillate around irrational number

Question

If $\alpha$ is an irrational number and $\frac{p_n}{q_n}$ for $n\geqslant 0$ are the convergents to $\alpha$.

Is there a way to show that the convergents alternate between being $<\alpha$ and $>\alpha$?

Answer 1

Suppose that $\frac{a}b<\frac{c}d$, where $a,b,c$, and $d$ are positive integers. For any $k\ell\in\Bbb Z^+$ we have

$$\frac{ka+\ell c}{kb+\ell d}-\frac{a}b=\frac{\ell(bc-ad)}{kb^2+\ell bd}>0$$

and

$$\frac{ka+\ell c}{kb+\ell d}-\frac{c}d=\frac{k(ad-bc)}{kbd+\ell d^2}<0\,,$$

so that

$$\frac{a}b<\frac{ka+\ell c}{kb+\ell d}<\frac{d}d\,.$$

In particular, it follows from this that $\frac{p_{k+2}}{q_{k+2}}$ always lies between $\frac{p_k}{q_k}$ and $\frac{p_{k+1}}{q_{k+1}}$. Use this to show that the even convergents are increasing, the odd convergents are decreasing, and all of the even convergents are less than any of the odd convergents.