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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.

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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 1 '13 at 6:18
    
$a_k \to 0$ since the given series converges. –  The Substitute Jul 1 '13 at 6:20
    
On page 1, Khinchin says we assume $a_1,a_2,\ldots$ are positive integers. –  The Substitute Jul 1 '13 at 6:22
    
@TheSubstitute For the purposes of this proof, there is no requirement that $a_n$ be an integer, only positive. –  Erick Wong Jul 1 '13 at 6:46
    
I agree. Whether or not $a_k$ is an integer, I don't see how $(*)$ can hold for any $l<k$. –  The Substitute Jul 1 '13 at 6:47

1 Answer 1

up vote 6 down vote accepted

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 $l<k$.

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4  
@ Erick Wong: Original Russian text is the same. –  Boris Novikov Jul 1 '13 at 9:37

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