# Adding integers to an infinite continued fraction expansion doesn't change the value?

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 $a_0$.

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+1: For showing what you did so far. –  Aryabhata Oct 29 '10 at 14:07
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). –  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!

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

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

$\vdots$

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

$\vdots$

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 $\lim_{n \to \infty} \beta_n = \xi$.

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Thank you for pointing out my errors, Moron. I did my best to correct them, and I think I've got it now. –  yunone Oct 30 '10 at 2:24
@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! –  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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+1: Now I agree with this :-) –  Aryabhata Oct 30 '10 at 18:19