I'm learning about continued fractions, and I've enjoyed them so far, but I'm unsure if I've done the following correctly. I have no real experience with analysis, so I'm not sure if my reasoning is formal enough, or correct. Any feedback would be appreciated.

Let $\xi$ be an irrational number with continued fraction expansion $\langle a_0,a_1,a_2,a_3\dots\rangle$. Let $b_1,b_2,b_3,\cdots$ be any sequence of positive integers, either finite or infinte. Prove that $\lim_{n\to\infty}\langle a_0,a_1,a_2,\dots,a_n,b_1,b_2,b_3\dots\rangle=\xi$.

I let $r_n=\langle a_0,a_1,\dots,a_n\rangle$ and $\xi'=\langle b_1,b_2,b_3,\dots\rangle$, and let $\beta_n=\langle a_0,a_1,a_2,a_3\dots,a_n,\xi'\rangle$. So*

$$\beta_n-r_n=\beta_n-\frac{h_n}{k_n}=\frac{\xi'h_n+h_{n-1}}{\xi'k_n+k_{n-1}}-\frac{h_n}{k_n}$$ $$=\frac{-(h_nk_{n-1}-h_{n-1}k_n)}{k_n(\xi'k_n+k_{n-1})}=\frac{(-1)^n}{k_n(\xi'k_n+k_{n-1})}$$

But ${k_n}$ is a positive increasing series, $\xi'$ is a positive real number, so as $n$ approaches $\infty$, the denominator goes to $\infty$ while the numerator alternates between $-1$ and $1$, so the fraction tends to $0$. Hence we have that $\lim_{n\to\infty}(\beta_n-r_n)=0$, so $$\lim_{n\to\infty}\beta_n=\lim_{n\to\infty}r_n=\lim_{n\to\infty}\langle a_0,a_1,a_2,\dots,a_n\rangle=\langle a_0,a_1,a_2\dots\rangle=\xi.$$ Hence $\lim_{n\to\infty}\langle a_0,a_1,a_2,\dots,a_n,b_1,b_2,b_3\dots\rangle=\xi$.

If this the correct route to go? As a small side question, how does it make sense to have integers $b_i$ at the end of this sequence, if the sequence is infinite? Thanks!

*If it's not well known, $\{h_n\}$ is the sequence defined by $h_{-2}=0,h_{-1}=1,h_i=a_ih_{i-1}+h_{i-2}$ and $\{k_n\}$ is defined as $k_{-2}=1,k_{-1}=-1, k_i=a_ik_{i-1}+k_{i-2}$, and $r_n=\langle a_0,a_1,\dots,a_n\rangle$, for any sequence of integers $a_0,a_1,a_2\dots$ all positive except perhaps $a_0$.

  • 1
    $\begingroup$ b_1, b_2, ... itself defines the continued fraction of some real b, so it suffices to prove the statement for a single real b at the end of the continued fraction. As for the side question, it's not integers at the end of an infinite sequence; it's a sequence of (integers at the end of a finite sequence). $\endgroup$ – Qiaochu Yuan Oct 29 '10 at 14:10

You are on the right track, but some statements do not make any sense at all.

1) "Since $\xi'$ is a sequence of positive integers" : This does not make any sense. $\xi'$ is a constant real number.

2) You are using confusing notation. $\beta = \langle a_0, a_1, \dots, a_n, \xi' \rangle$. This varies with $n$, so you need to talk about $\beta_n$.

3) "Hence we have $\lim_{n \to \infty} (\xi' -r_n) = 0$" is not right. This implies that any two real numbers are equal (think about it)! You need to consider $\lim_{n \to \infty} \beta_n$.

As to you side question, it is not one sequence.

It is a sequence of sequences!

$a_0, a_1, b_1, b_2, b_3, \dots$

$a_0, a_1, a_2, b_1, b_2, b_3, \dots$


$a_0, a_1, a_2, \dots, a_n, b_1, b_2, b_3 \dots$


Notice that in each sequence, we only had a finite number of the $a_i$.

Each sequence corresponds to a real number (the $\beta_n$ above).

Thus we get a new sequence

$\beta_1, \beta_2, \dots$

And the problem is asking you to prove that $\lim_{n \to \infty} \beta_n = \xi$.

  • $\begingroup$ Thank you for pointing out my errors, Moron. I did my best to correct them, and I think I've got it now. $\endgroup$ – yunone Oct 30 '10 at 2:24
  • $\begingroup$ @CRom: Yes, you also need to mention that $\beta_n = \langle a_0, a_1, \dots, a_n, b_1, b_2, \dots \rangle $. Other that, it looks fine to me. btw, nicely done! $\endgroup$ – Aryabhata Oct 30 '10 at 15:34

Your work looks good to me.

It does make sense to have the $b_n$ at the end of the sequence as each stage of the sequence is finite. This is an important point. At each stage, the list of $a$'s is finite, so it makes sense to have more numbers (of any sort) at the end. But as you let $n \rightarrow \inf $ they get pushed farther and farther out. The way continued fractions work, they have less and less impact, as you showed. So, "in the limit", they don't matter.


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