How can I show that if $x=[a_0;a_1,a_2,\dots]$, then |x-C_k|<1/a_k^{\text{}}q_k^2 using the facts that
$\begin{align} C_k-C_{k-1}&=\frac{(-1)^{k-1}}{q_kq_{k-1}}\text{, and}\\ |C_k-C_{k-1}|&\leqslant\frac{1}{q_kq_{k-1}}? \end{align}$
This is what I have done:
|C_k-C_{k-1}|\leqslant\frac{1}{q_kq_{k-1}}=\frac{1}{(a_kq_{k-1}+q_{k-2})q_{k-1}}=\frac{1}{a_k^{\text{}}q_{k-1}^2+q_{k-1}q_{k-2}}<\frac{1}{a_k^{\text{}}q_{k-1}^2}.
However, since $(q_k)_{k\in\mathbb{N}\cup\{0\}}$ is an increasing sequence,
$\frac{1}{a_k^{\text{}}q_{k-1}^2}>\frac{1}{a_k^{\text{}}q_{k}^2}.$
Therefore, I cannot assume that the inequality would still hold if I increased the index. Moreover, I am not sure about the relationship between $|x-C_k|$ and $|C_k-C_{k-1}|$.