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I'm trying to prove the following result:

Let $n$ be a fixed natural number and $a_1, a_2, \cdots, a_n$ and $b_1, b_2, \cdots, b_n$ fiexed real values.

Then given $\varepsilon>0$, there exits a $\delta>0$ such that if $\left| r-s \right| < \delta$ then $$ \left| \left( a_1, b_2|a_2, \cdots, b_n|a_n,r \right) - \left( a_1, b_2|a_2, \cdots, b_n|a_n,s \right) \right| < \varepsilon$$

where

$$\left( a_1, b_2|a_2, \cdots, b_n|a_n \right)=a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cfrac{b_4}{\ddots +\cfrac{b_n}{a_n}}}}$$

$$\left( a_1, b_2|a_2, \cdots, b_n|a_n, \xi \right)=a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cfrac{b_4}{\ddots +\cfrac{b_n}{a_n+\xi}}}}$$

I also want to prove that, given $\varepsilon>0$ there exits a $\delta>0$ such that if $$ \left| \left( a_1, b_2|a_2, \cdots, b_n|a_n,r \right) - \left( a_1, b_2|a_2, \cdots, b_n|a_n,s \right) \right| < \delta$$ then $\left| r-s \right| < \varepsilon$

I've proven that if

$$(a_1,b_2|a_2, \cdots, b_n|a_n)=\dfrac{p_n}{q_n}$$

then if we denote $p_0=1$, $p_1=a_1$, $q_1=1$ and $q_2=a_2$ for $n \geq 2$

$$p_n=a_np_{n-1}+b_np_{n-2}$$ $$q_n=a_nq_{n-1}+b_nq_{n-2}$$

Also, I've got that for $n \geq 2$

$$\left( a_1, b_2|a_2, \cdots, b_n|a_n, \xi \right)=\dfrac{p_n+\xi p_{n-1}}{q_n+\xi q_{n-1}}$$

And, from the previous result, I've expanded the difference

\begin{align} {\left| \left( a_1, b_2|a_2, \cdots, b_n|a_n,r \right) - \left( a_1, b_2|a_2, \cdots, b_n|a_n,s \right) \right|=\left| \dfrac{p_n+r p_{n-1}}{q_n+r q_{n-1}}-\dfrac{p_n+s p_{n-1}}{q_n+s q_{n-1}} \right|=\\ =|r-s| \left| \dfrac{p_nq_{n-1} - p_{n-1}q_n}{(q_n+rq_{n-1})(q_n+sq_{n-1})} \right|\\} \end{align}

Now, I don't know how to bound the last fraction, I guess it should be less than $1$, but I don't know how to continue. Any ideas?

Eparoh
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  • Why do you think that the "iff" result is true? Have you tried to show it for $n=2$? – Somos Jan 08 '20 at 22:11
  • Mmmm, I think I've stated in a wrong way the result I want to achieve. What I was looking is to show that you can make the difference of the two continued fractions I propose as small as you want by making the difference of $r$ and $s$ as small as needed, and vice versa. – Eparoh Jan 09 '20 at 20:19
  • Please correct your question accordingly. I ask again. Have you tried to show it for $n=2$? – Somos Jan 09 '20 at 21:06
  • I've edited the question. Yes, I've tried to prove it for $n=2$ and although I can't prove it rigorously, what you get after simplifying the fraction is $\left| \frac{b_2}{a_2+r}- \frac{b_2}{a_2+s} \right|$ so if $r$ and $s$ are really close, then both numbers in the absolute value would also be close to each other, but mixing sums, products and quotients are a mess, and I cannot bound the expression to fully justify it. And for the reciprocal, if the two fractions are close, then as they have the same numerator, the denominators must be close and so are $s$ and $r$. – Eparoh Jan 10 '20 at 22:44
  • You need to clarify the quantification order. Are you trying to show that “for all $n, a_1, \dotsc, a_n, b_1, \dotsc, b_n$, given $\epsilon > 0$, there exists $\delta > 0$ such that for all $r, s$ with $|r - s| < \delta$, we have (inequality)”? Or are you trying to show that “given $\epsilon > 0$, there exists $\delta > 0$ such that for all $r, s$ with $|r - s| < \delta$, we have for all $n, a_1, \dotsc, a_n, b_1, \dotsc, b_n$, (inequality)”? – Anders Kaseorg Jan 11 '20 at 04:26
  • @Eparoh I concur with Anders Kaseorg, and I would add this : if you allow $b_2$ to be any real, you can always find a $b_2$ such that $\bigg|\frac{b_2}{a_2+r}-\frac{b_2}{a_2+s}\bigg|$ is $1$, so it will not be arbitrarily small. – Ewan Delanoy Jan 11 '20 at 14:55
  • I've edited it again, $a_1, \cdots, a_n$ and $b_1, \cdots, b_n$ are fixed values. What I want to show is that when you fixed that $a$'s and $b$'s you can make the difference of the continued fraction I propose as small as you want by choosing $r$ and $s$ as closed as needed. And viceversa. – Eparoh Jan 13 '20 at 12:39

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