Mistake in Khinchin's "Continued Fractions"

Question

I am reading Khinchin's Continued Fractions page 10.

$\lbrack a_1;a_2,a_3\ldots\rbrack$ is a continued fraction and $q_k$ is given by $q_k=a_kq_{k-1}+q_{k-2}$. Suppose $\sum_{n=1}^{\infty}a_n$ converges so that there is a $k_0$ for which $k\ge k_0$ implies $a_k<1$.

Khinchin says "for $k\ge k_0$ we have $$q_k<\frac{q_l}{1-a_k} \, (*)$$

where $l<k$."

Problem: Since $a_k\to0, $ the right hand of $(*)$ side tends to $q_l$, which is fixed. However the left side tends to infinity. Since $q_k$ is a growing sequence, this cannot happen.

Why does $a_k\rightarrow 0$? In the usual definition, they are arbitrary integers. In the generalized definition, they can be any complex numbers but again, there's no requirement that they go to 0. — Alex R., Jul 01 '13 at 06:18
On page 1, Khinchin says we assume $a_1,a_2,\ldots$ are positive integers. — The Substitute, Jul 01 '13 at 06:22
@TheSubstitute For the purposes of this proof, there is no requirement that $a_n$ be an integer, only positive. — Erick Wong, Jul 01 '13 at 06:46
I agree. Whether or not $a_k$ is an integer, I don't see how $(*)$ can hold for any $l<k$. — The Substitute, Jul 01 '13 at 06:47
@TheSubstitute In the case that $a_k < 1$, the sequence ${q_n}$ does not necessarily grow to infinity (imagine the $a_k$'s decrease very rapidly to $0$). — Erick Wong, Jul 01 '13 at 06:50
@ErickWong, what if $l=1$? Then we'd have that $q_n$ is actually decreasing. — The Substitute, Jul 01 '13 at 06:52
@TheSubstitute I understand your question now, and have an answer. — Erick Wong, Jul 01 '13 at 06:59

score 6 · Accepted Answer · answered Jul 01 '13 at 07:04

This looks like a bit of a looseness, perhaps in translation. The quoted inequality is not meant to be universally quantified in $l$. If you read the two paragraphs leading up to it, they show that either $q_k < q_{k-1} / (1-a_k)$ or $q_k < q_{k-2} / (1-a_k)$.

Thus the author simply means to say that $(*)$ holds for some $l < k$. If you read the inequality that follows you'll see it is applying this inductively to connect $q_k$ further back to a term earlier than $q_{k_0}$. This repeated application wouldn't be necessary if $(*)$ were true for all $l<k$.

@ Erick Wong: Original Russian text is the same. – Boris Novikov Jul 01 '13 at 09:37 — Boris Novikov, Jul 01 '13 at 09:37

score 0 · Answer 2 · answered Feb 03 '17 at 19:59

Khinchin's text is indeed a bit confusing -- same goes for Russian original as well. On one page he says he is going to consider only integer values of $a_k$ but in this theorem it is kind of obvious that the proof is valid for any sequence of positive real numbers $a_k$. A meta-explanation of why we should (at least for this theorem) think of $a_k$ as real is clear -- if they are positive integers (as is the case with standard continued fractions) then $\sum_{k=1}^\infty a_k$ always diverges and that condition is basically pointless.

Mistake in Khinchin's "Continued Fractions"

2 Answers2