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In an article it is used the fact that $$q_n\left|q_n\alpha-p_n\right|(a_{n+1}+1)>1$$ where $\alpha=[a_0;a_1,\ldots]$ is an irrational number and $q_i$ is the series of the best approximation denominator of $\alpha$, ie it is the sequence of the denominators of the sequence of the convergents of $\alpha$. I am not able to prove it, any idea?

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  • $\begingroup$ Potentially related inequality: $q_{n+1}=a_{n+1}q_n+q_{n-1}< q_n(a_{n+1}+1)$ $\endgroup$ – Thomas Andrews Aug 16 '14 at 13:39
  • $\begingroup$ I am not able to obtain the thesis using this :( $\endgroup$ – Nisba Aug 16 '14 at 13:48
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    $\begingroup$ It may can be useful the fact (wikipedia) $|q_n\alpha-p_n|>\frac{1}{q_{n+1}+q_n}$ $\endgroup$ – Nisba Aug 16 '14 at 14:21
  • $\begingroup$ Yeah, that's an essentially weaker inequality than the one in the problem, since $$\frac{1}{q_{n+1} +q_n}<\frac{1}{(a_{n+1}+1)q_n}$$ $\endgroup$ – Thomas Andrews Aug 16 '14 at 14:43
  • $\begingroup$ With the inequality you gave and using wikipedia's one, I have proved a weaker version of what I want to prove: instead of $(a_{n+1}+1)$ I have $(a_{n+1}+2)$ $\endgroup$ – Nisba Aug 16 '14 at 14:46
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Counterexample:

$$\alpha=\frac{1+\sqrt{5}}{2}, p_n=3,q_n=2, a_{n+1}=1$$

Another counterexample:

$$\alpha=\frac{1+\sqrt{3}}{2}, p_n=4, q_n=3, a_{n+1}=2$$

I can show it is true when $a_n\geq 2$.

Your inequality is equivalent to:

$$\left|\alpha-\frac{p_n}{q_n}\right|>\frac{1}{(a_{n+1}+1)q_n^2}$$

Now assume the opposite, so that:

$$\left|\alpha-\frac{p_n}{q_n}\right|<\frac{1}{(a_{n+1}+1)q_n^2}$$

From $$\frac{1}{q_{n+1}q_n}=\left|\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n}\right|= \left|\frac{p_{n+1}}{q_{n+1}}-\alpha\right|+\left|\alpha-\frac{p_n}{q_n}\right| $$ we see:

$$\left|\frac{p_{n+1}}{q_{n+1}}-\alpha\right|=\frac{1}{q_{n+1}q_n}-\left|\frac{p_{n}}{q_{n}}-\alpha\right|>\frac{1}{q_{n+1}q_n}-\frac{1}{(a_{n+1}+1)q_n^2}=\frac{(a_{n+1}+1)q_n-q_{n+1}}{q_{n+1}q_n^2}$$

But $q_{n+1}=a_{n+1}q_n+q_{n-1}$, so:

$$\left|\frac{p_{n+1}}{q_{n+1}}-\alpha\right|> \frac{q_n-q_{n-1}}{q_{n+1}q_n^2}$$

When we can show that $\frac{q_n-q_{n-1}}{q_n}\geq\frac{1}{2}$, we'd be done, because this would mean that:

$$\left|\frac{p_{n+1}}{q_{n+1}}-\alpha\right|\geq\frac{1}{2q_nq_{n+1}}$$

But $\frac{q_n-q_{n-1}}{q_n}>\frac 12$ is true when $a_n\geq 1$, since:

$$\frac{q_n-q_{n-1}}{q_n} = \frac{(a_n-1)q_{n-1}+q_{n-2}}{a_nq_{n-1}+q_{n-2}}\geq\frac{a_{n}-1}{a_n}\geq \frac{1}{2}$$

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  • $\begingroup$ Thank you very much! For the case $a_n=1$ I have proved (it is true in general) a weaker version: $q_n\left|q_n\alpha-p_n\right|(a_{n+1 + \frac{1}{a_n}}+1)>1$. The proof follows easily from $|q_n-\alpha p_n|>\frac{1}{q_n+q_{n+1}}$ and using the fact that $q_n>a_n q_{n-1}$ for $n>1$. $\endgroup$ – Nisba Aug 16 '14 at 16:20
  • $\begingroup$ I doubt you mean $a_{n+1+1/a_n}+1$ which only makes sense when $a_n=1$, but perhaps $a_{n+1}+\frac{1}{a_n}+1$? $\endgroup$ – Thomas Andrews Aug 16 '14 at 16:25
  • $\begingroup$ It's also easy to prove that $$q_n\left|q_n\alpha-p_n\right|(a_{n+1}+1)>\frac{1}{2}$$ $\endgroup$ – Thomas Andrews Aug 16 '14 at 16:26
  • $\begingroup$ Yes I meant exactly what you wrote! $\endgroup$ – Nisba Aug 18 '14 at 16:52

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