Odd Convergents of Simple Continued Fraction are Strictly Decreasing

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Theorem

Let $n \in \N \cup \set \infty$ be an extended natural number.

Let $\sqbrk {a_0, a_1, \ldots}$ be a simple continued fraction in $\R$ of length $n$.

Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.

Let $\sequence {C_0, C_1, \ldots}$ be its sequence of convergents.


The odd convergents satisfy $C_1 > C_3 > C_5 > \cdots$


Proof

Let $ k \ge 3$ be an odd integer.

From Difference between Adjacent Convergents But One of Simple Continued Fraction:

$C_k - C_{k - 2} = \dfrac {\paren {-1}^k a_k} {q_k q_{k - 2} } = \dfrac {-a_k} {q_k q_{k - 2} }$

By definition of simple continued fraction, $a_k > 0$.

By Convergents of Simple Continued Fraction are Rationals in Canonical Form, $q_k > 0$ and $q_{k - 2} > 0$.

Thus $C_k < C_{k - 2}$.

$\blacksquare$


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