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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 $|C_k-C_{k-1}|$.

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1 Answer 1

up vote 1 down vote accepted

Are you allowed to use the fact that $x$ is always between $C_k$ and $C_{k+1}$? If so, you have $|x-C_k|\lt|C_{k+1}-C_k|$ and then your argument gives you $|x-C_k|\lt1/(a_{k+1}q_k^2)$. This is not exactly what you asked for, but I'm not convinced that the inequality you asked for is correct.

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I also had the same feeling about this inequality. I will comment here again when I check whether this is a typo or not. Thanks! –  Josué Molina Apr 19 '12 at 3:43
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