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$\newcommand{\K}{\operatorname{\large\mathcal{K}}}$I am asking about a very common practice in proofs, that I see online, concerning continued fractions. There is an implicit assumption which I’d like to examine: a concrete, concise question can be found in the middle section, but I think a lot of context is necessary. Here is what I mean: sometimes, when dealing with continued fractions, we begin with the answer, and then recursively expand the answer, ad infinitum, implicitly assuming that the limit will be the same.

I occasionally edit this post to add new thoughts, and sufficient conditions (a necessary one would be very well-received!) in the middle section.

To clarify, here is a trivial example:

$$\begin{align}1&=\cfrac{2}{3-1}\\&=\cfrac{2}{3-\cfrac{2}{3-1}}\\&=\cfrac{2}{3-\cfrac{2}{3-\cfrac{2}{3-1}}}\\&\overset{?}{=}\cdots\\&=\cfrac{2}{3-\cfrac{2}{3-\cfrac{2}{3-\cfrac{2}{3-\ddots}}}}\end{align}$$

We started with $1$, then expressed it as a ratio $\frac{b_1}{a_1+\alpha_1}$. We then expanded $\alpha_1$ as a ratio $\frac{b_2}{a_2+\alpha_2}$... et cetera. We can show that (Gauss Kettenbrücher K-notation, similar to $\sum,\prod$ notation) $\K_{n=1}^\infty\frac{b_n}{a_n}$ is convergent, and because we obtained this continued fraction from a sequence of expressions, all equal to $1$, we have reason to believe that the limiting fraction is also equal to $1$. However, in so doing we lose the helper $\alpha_n$ terms, so I feel, a priori, we cannot be certain that the limit is actually $1$. It is, as it happens... the convergents are $-\K_{n=1}^m\frac{-2}{3}=1-\frac{1}{2^{m+1}-1}$, which obviously tend to $1$.

But I ask about the general procedure. For example, Wikipedia’s article on Gauss’ continued fraction uses recurring expansions, just like how we expanded “$1$” above, to derive continued fractions for ratios of hypergeometric functions. I like this derivation, but there is always the concern that, even if the continued fraction is convergent, the limiting value might not be the value we started with!

A simple example to demonstrate this concern: in the above expansion of $1$, we could just as easily write: $$\begin{align}2&=\frac{2}{3-2}\\&=\cfrac{2}{3-\cfrac{2}{3-2}}\\&=\cdots\\&\overset{!?}{=}\cfrac{2}{3-\cfrac{2}{3-\cfrac{2}{3-\ddots}}}\end{align}$$But we already know that that converges to $1$, not to $2$!


The core question.

My analysis, and a concrete setup:

Let $L$ be any nonzero real number. Suppose we find a sequence of nonzero reals $(a_k),(b_k),(\alpha_k)$ with the property that, for all $n\in\Bbb N$: $$L=\K_{k=1}^n\frac{b_k}{a’_k}$$Where $a’_k=a_k$ except for $a’_n:=\alpha_n$. Suppose further that: $$\lim_{n\to\infty}\K_{k=1}^n\frac{b_k}{a_k}=L’\in\Bbb R\setminus\{0\}$$That is, the continued fraction converges. We want to know if $L’=L$. If $p_n,q_n$ are the numerators and denominators of the $n$th convergent $\K_{k=1}^n\frac{b_k}{a_k}$ (with no simplification!) then we know that (writing $\alpha_n=\alpha_n/1$): $$L=\frac{p_n+\alpha_np_{n-1}}{q_n+\alpha_nq_{n-1}}$$For every $n$. The difference $|L-L’|$ is equal to: $$\left|\frac{q_n\left(\frac{p_n}{q_n}-L’\right)+\alpha_nq_{n-1}\left(\frac{p_{n-1}}{q_{n-1}}-L’\right)}{q_n+\alpha_nq_{n-1}}\right|=\left|\left(\frac{p_n}{q_n}-L’\right)+\cfrac{1}{1+\cfrac{q_n}{\alpha_nq_{n-1}}}\left(\frac{p_{n-1}}{q_{n-1}}-\frac{p_n}{q_n}\right)\right|$$For every $n$. Thus I can hope to send $n\to\infty$ - so long as $\frac{q_n}{\alpha_nq_{n-1}}$ is bounded away from $-1$, it is easy to see all terms tend to zero, whence $L=L’$.

However, there’s the rub - can we necessarily say that there is some $\delta>0$, that for any $n\in\Bbb N$, $\left|\frac{q_n}{\alpha_nq_{n-1}}+1\right|>\delta$? One quick observation is that, if all $(a_k),(b_k),(\alpha_k)$ are positive, then the answer is a definite: ‘yes’. Another observation, inspired by the partial answer below, is that I really only need to find a subsequence $(n_k)$ along which $\frac{q_{n_k}}{\alpha_{n_k}q_{n_k-1}}$ is bounded away from $-1$.

Note: if no such subsequence can be found, that actually implies the stronger assertion that: $\lim_{n\to\infty}\alpha_n\frac{q_{n-1}}{q_n}=-1$ as a strong limit.

Furthermore, we see that, in order for $L$ to be finite and well defined, if $q_n/\alpha_n q_{n-1}$ comes arbitrarily close to $-1$, then $\frac{\alpha_n p_{n-1}}{q_n}$ comes arbitrarily close to $-L’$. I’ve tried using that to obtain a contradiction, but I’ve had no such luck yet. Furthermore, $\alpha_np_{n-1}/p_n$ would also tend to $-1$.

A new observation: if $\alpha_n\to0$, as happens in the case of Lambert’s tangent continued fraction, and if one can be sure that $|q_n|\ge|q_{n-1}|$ for large $n$, or at least along a subsequence (more generally, we only need to assume the ratio $q_{n-1}/q_n$ is bounded) - which is often the case - then $\alpha_n q_{n-1}/q_n$ tends to $0$, and is in particular bounded away from $-1$. This is not a necessary condition, since my trivial example involving $1=3/2-\cdots$ has $\alpha_n=-1$ for all $n$. If the $\alpha_n$ are constant, then so long as we can say $|q_n|$ is sufficiently larger than $|q_{n-1}|$ (eventually, perhaps along a subsequence) then it will also be possible to bound away from $-1$ (more generally, the condition would be the unwieldy: “$q_{n-1}/q_n$ is bounded away from the constant $-1/\alpha_n$”). See the counterexample expansion of $2$: $-1/\alpha=\frac{1}{2}$ in this case, and $1/2$ is precisely the limiting ratio of $\frac{q_{n-1}}{q_n}$, explaining why convergence fails.

I don’t however think that this is anywhere near a complete, or usefully general, list. There is surely more to this story.

My question: can we say more about the general case? I’d be happy to see relatively simple, easy-to-apply conditions, other than the conditions I’ve already supplied. As I have shown, and more recently G. Edgar has shown, counterexamples do exist.


A less trivial, and more motivating, example: through some series manipulations and divisions of the Maclaurin expansions for $\sin,\cos$, we can find: $$\tan x=\cfrac{x}{1-\cfrac{x^2}{3-\cfrac{x^2}{P_1/P_2}}}$$Where: $$P_m:=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2(n+m)+1)!}\prod_{k=1}^m2(n+k)$$It can be shown that, using identical manipulations: $$\frac{P_m}{P_{m+1}}=(2m+3)-\frac{x^2}{P_{m+1}/P_{m+2}}$$So, inductively, a continued fraction for $\tan x$ is born: $$\tan x=\cfrac{x}{1-\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-\ddots}}}}$$This continued fraction can be shown to converge - but does it necessarily converge to $\tan x$? The concern being, again, that we have lost the helper $\alpha_m:=-x^2/(P_m/P_{m+1})$ terms.

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    $\begingroup$ I'm surprised this hasn't been asked here before. $\endgroup$ Aug 23, 2022 at 13:37
  • $\begingroup$ @moleculesdumal Indeed. It is implied all over the Internet, but seems to be a rather important point. $\endgroup$
    – FShrike
    Aug 23, 2022 at 13:38
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    $\begingroup$ I mean usually you start with a convergent series expansion and then you come up with a continued fraction expansion whose convergents are equal to the partial sums of the series. Euler did this very nicely here $\endgroup$
    – dezdichado
    Aug 23, 2022 at 17:08
  • $\begingroup$ @dezdichado Quite. I am asking about the other method, described above, which I see very often. It can be used to prove otherwise very difficult results, such as the many of Gauss’ continued fractions. Euler’s method is nice, but not entirely general. For example, I’d be astonished if you could use it to derive the above tangent fraction! $\endgroup$
    – FShrike
    Aug 23, 2022 at 17:11
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    $\begingroup$ The starting value is $\tan x$. One breaks it down as I describe, using ratios of $P_m/P_{m+1}$ as I describe, with the continued fraction equal to $\tan x$ for every $m$. $\endgroup$
    – FShrike
    Aug 23, 2022 at 17:38

2 Answers 2

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Interesting question. Of course we can solve linear equations to get rational numbers $b_1,b_2,b_3,\dots$ with $$ 3 = \cfrac{1}{1+b_1} = \cfrac{1}{1+\cfrac{1}{1+b_2}} = \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+b_3}}}=\dots $$ namely: $$ b_1 = -\frac{2}{3},\quad b_2 = -\frac{5}{2},\quad b_3 = -\frac{7}{5},\quad\text{and so on.} $$ but the infinite continued fraction $$ \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ddots}}} $$ converges to $(\sqrt5 - 1)/2$, not $3$.

Indeed, with any infinite continued fraction and any alleged value, we can solve linear equations to get $b_1,b_2,b_3,\dots$ that work. (There are a few exceptional cases where the linear equation for $b_n$ requires dividing by zero.)


FShrike commented, "It is implied all over the Internet".

See HERE where "J. M. ain't a mathematician" has done this, but quoted a result of Pincherle to claim convergence to the right value in that case.

Maybe that result of Pincherle is what our OP needs!

Image from Introduction to Difference Equations by S. Elaydi:

S Elaydi

From the references, the paper of Pincherle:
references

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  • $\begingroup$ Another nice counterexample. $\endgroup$
    – FShrike
    Aug 28, 2022 at 11:05
  • $\begingroup$ Aha! I will look into this Pincherle result. $\endgroup$
    – FShrike
    Aug 28, 2022 at 11:17
  • $\begingroup$ It’s not immediately obvious in what sense a solution is “minimal”. I don’t know if I’d need to scroll through the whole book to find this. Does it just want $\varphi(n)$ to attain the smallest possible values? $\endgroup$
    – FShrike
    Aug 28, 2022 at 11:20
  • $\begingroup$ It seems section 9.6 "Minimal Solutions, Continued Fractions, and Orthogonal Polynomials" is the place to look. But page 422 is not in the free preview. Also the references are not in the free preview to help us find the paper of Pincherle. $\endgroup$
    – GEdgar
    Aug 28, 2022 at 11:28
  • $\begingroup$ Found the reference for Pincherle. Added. $\endgroup$
    – GEdgar
    Aug 28, 2022 at 11:43
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Think this is works...

You showed the positive sign case. We'll show the negative sign case converges with simple conditions. I don't see a reason why the arbirary sign case couldn't converge to any number when $\alpha_n$ are removed, although the arbitrary case is not common.

Suppose you have extended $$ L = \frac{b_1}{a_1-} \frac{b_2}{a_2-}\cdots \frac{b_n}{a_n - \alpha_n} $$ where $a_k,b_k,\alpha_n >0$ such that at each step $a_k - \alpha _k > 0$. Denote the RHS by $A_n$. Denote $L_n = \frac{b_1}{a_1-} \frac{b_2}{a_2-}\cdots \frac{b_n}{a_n}$. Now
$$ \alpha_{n-1} = \frac{b_n}{a_n - \alpha_n} > \frac{b_n}{a_n} $$ By induction on $k$ and hypothesis of the construction $$ a_{n-k-1} > \frac{b_{n-k}}{a_{n-k}-} \cdots \frac{b_n}{a_n - \alpha_n} > \frac{b_{n-k}}{a_{n-k}-} \cdots \frac{b_n}{a_n} > 0 $$ for all $0 \leq k < n$. This shows $L = A_n > L_n > 0$. Similarly one can argue $L_n < L_{n+1}$ by replacing $\alpha_n$ with $b_{n+1}/a_{n+1}$. So $L_n$ is increasing, bounded above, and hence converges.

We will construct $p_n,q_n > 0$ such that $\frac{p_n}{q_n} = L_n$. Define $p_1,q_1,p_2,q_2,p_3,q_3$ explicitly and \begin{align} p_n &:= a_np_{n-1} - b_n p_{n-2} \\ q_n &:= a_n q_{n-1} - b_n q_{n-2} \end{align} Suppose by induction $p_n/q_n = L_n$. The base case is straightforward. Replace $a_n$ with $a_n - a_{n+1}/b_{n+1}$. \begin{align} \left(a_n - \frac{b_{n+1}}{a_{n+1}}\right) p_{n-1} - b_n p_{n-2} &= p_n - \frac{b_{n+1}}{a_{n+1}} p_{n-1} \\ \left(a_n - \frac{b_{n+1}}{a_{n+1}}\right) q_{n-1} - b_n q_{n-2} &= q_n - \frac{b_{n+1}}{a_{n+1}} q_{n-1} \end{align} The bottom is not $0$, since otherwise $L_{n+1}$ is undefined by first dividing and then replacing. It follows $p_{n+1}/q_{n+1} = \mathcal{K}_{k=1}^{n+1} b_k/a_k$ and we're done. With our recursive identities its simple to show $$ L_{n+1}-L_n = \frac{b_{n+1}q_{n-1}}{q_{n+1}}\left(L_n - L_{n-1}\right) $$ Replacing $a_{n+1}$ by $a_{n+1} - \alpha_{n+1}$ in the previous formula gives $$ L - L_n = \frac{b_{n+1}q_{n-1}}{q_n - \alpha_{n+1}q_{n-1}} \left(L_n - L_{n-1} \right) $$ Since $L_n$ converges, the only way $L_n \nrightarrow L$ is if $b_{n+1}/|q_n/q_{n-1}-\alpha_{n+1}| \rightarrow \infty$. Suppose then $q_n/q_{n-1} \rightarrow \alpha_{n+1}$. Dividing by $q_{n-1}$ and letting $n \rightarrow \infty$ in the recursive formula for $q_n$

$$ \alpha_{n+1} - a_n - b_n/\alpha_n \rightarrow 0 $$ Thus if one can find a subsequence $b_{n_m}$ that is bounded and $\alpha_{n_m+1} - a_{n_m} - b_{n_m}/\alpha_{n_m} \nrightarrow 0 $ then $$ L_n \rightarrow L $$ which seems it'd be commonly possible to find (if not, one might still improve upon this by finding another way to show $b_{n+1}/|q_n/q_{n-1}-\alpha_{n+1}| \nrightarrow \infty$).

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  • $\begingroup$ Thank you for your thoughts so far. I think: $$\mathcal{K}_{k=1}^n(-1)^{k+1}\frac{b_k}{a_k}=\frac{b_1}{a_1-}\frac{b_2}{a_2\color{red}{+}}\frac{b_3}{a_3-}\cdots$$Which is different in nature to $L$. $\endgroup$
    – FShrike
    Aug 23, 2022 at 20:44
  • $\begingroup$ Corrected, yea just wanted the partial fraction. $\endgroup$ Aug 23, 2022 at 20:46
  • $\begingroup$ I mean, one has $-,+,-,+...$ and the other has $-,-,-,-,...$ $\endgroup$
    – FShrike
    Aug 23, 2022 at 20:51
  • $\begingroup$ If your question is for arbitrary signs I think this would go well beyond the +,+,+,... and -,-,-,-,... cases which are typically the continued fractions seen in literature. One may be able to do the +,-,+,-,... case with work, but if the signs are arbitrary I almost feel you can get anything. $\endgroup$ Aug 23, 2022 at 20:56
  • $\begingroup$ You misunderstand my point. Within your own example, you are considering conflicting fractions, where one alternates in sign and the other does not, so that makes your post harder to follow. I’m also now sure what you mean by: “retracting back”, here $\endgroup$
    – FShrike
    Aug 23, 2022 at 21:02

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