Continued fractions: convergence of fraction expansion

Using the notation $[a_0,a_1,a_2,...]$ for the continued fraction

$$a_0 + \dfrac{1}{a_1 + \dfrac{1}{a_2 + \dfrac{1}{...}}}$$

where $a_0 ∈ \mathbb Z$ and $a_i ∈ \mathbb N$ for all $i>0$

Question: Consider the continued fraction expansion $$[1,0,1,1,2,1,1,4,1,1,6,1,...,1,2n,1,...]$$

and let $p_i$ and $q_i$ denote the numerators and denominators (respectively) of its convergents. Prove that

$$p_{3n} = p_{3n-1} + p_{3n-2}, q_{3n} = q_{3n-1}+q_{3n-2}$$ $$p_{3n+1} = 2np_{3n} + p_{3n−1}, q_{3n+1} = 2nq_{3n} +q_{3n−1}$$ $$p_{3n+2} = p_{3n+1} + p_{3n}, q_{3n+2} = q_{3n+1} +q_{3n}$$

for $n=1,2,....$

As I understand it, the numerators are $[1,0,1,1,2,1,1,4,1,1,6,1,...,1,2n,1,...]$ while the denominator's are $[1,1,...]$. But I do not know what exactly is meant with "of its convergence".

I am also uncertain how to proceed with proving the above mentioned questions, but I do believe it relates to the "fundamental recurrence formulas" as I read a bit about in the following wikipedia article: https://en.wikipedia.org/wiki/Generalized_continued_fraction

Merry Christmas and happy new year.

• The convergents of a continued fraction are like partial sums of an infinite sum; to borrow your notation, the convergents of your continued fraction would be $a_0,a_0+\cfrac1{a_1},a_0+\cfrac1{a_1+\cfrac1{a_2}},\dots$. – J. M. is a poor mathematician Dec 28 '16 at 15:44
• I found the following document about continued fractions. – Sirmimer Dec 28 '16 at 16:10
• I found the following document about continued fractions: math.illinois.edu/~ajh/453.spring11/nt-notes6.pdf I still need to understand exactly how to apply these and where they ordinate from, but do you think I understand the assignment correctly, by saying this is what i need to understand and apply? – Sirmimer Dec 28 '16 at 16:11
• If the usual three-term recurrence for the numerators and denominators of the $n$-th convergent is in that document, then yes. – J. M. is a poor mathematician Dec 28 '16 at 16:14

It appears you don't know how to take the sequence of $a_i$ and produce the sequence of ordinary fractions called "convergents," often written $\frac{p_i}{q_i}.$ I have written out a few examples, sometimes for finite CF, sometimes for infinite.

Oh, the traditional way to start things is with $$\frac{p_0}{q_0} = \frac{a_0}{1}.$$ In order for all the formulas to work out, I draw in two pretend convergents at the left edge, giving $$\frac{p_{-2}}{q_{-2}} = \frac{0}{1},$$ $$\frac{p_{-1}}{q_{-1}} = \frac{1}{0},$$ $$\frac{p_0}{q_0} = \frac{a_0}{1}.$$ Note that the $1/0$ is fake, it is just a convenience, not a genuine fraction.

You have written the sequence of $a_i$ incorrectly, there may not be a zero as $a_1.$ Here is one correct possibility. If you change $a_0,$ all the denominators stay the same, each numerator changes by a fixed multiple of the relevant denominator.

I can recommend Hardy and Wright, chapter 10.

$$\begin{array}{ccccccccccccccccccccccc} & & 1 & & 1 & & 2 & & 1 & & 1 & & 4 & & 1 & & 1 & & 6 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{1}{1} & & \frac{2}{1} & & \frac{5}{3} & & \frac{7}{4} & & \frac{12}{7} & & \frac{55}{32} & & \frac{67}{39} & & \frac{122}{71} & & \frac{799}{465} \\ \end{array}$$

Solving Diophantine Equation - odd Periods

Problem in a Pell equation

Find a number $x \in \mathbb{Z}_{355213}$ such that $x = 2 \pmod {71}$ and $x = 13 \pmod {5003}$.

Is there a simpler way to find an inverse of a congruence?

Here are the basic rules from wikipedia. They use $h_n/k_n$ rather than $p_n/ q_n,$ but they keep the $a_n.$

• I believe you are correct. I have also read that the convergence of a infinite continued fraction as similar or exactly as you wrote above. – Sirmimer Dec 28 '16 at 18:00
• I will look at it in a bit and give feedback as to if i understand it or not. Thank you for the examples (links) above. – Sirmimer Dec 28 '16 at 18:00