Mistake in Khinchin's "Continued Fractions" 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.
 A: 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 $l<k$.
A: 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.
