Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent

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.

Every even convergent is strictly smaller than every odd convergent.

Proof
Let $k \ge 1$.

From Denominators of Simple Continued Fraction are Strictly Positive, $q_k q_{k - 1}>0$.

From Difference between Adjacent Convergents of Simple Continued Fraction:
 * $C_k - C_{k - 1} = \dfrac {\paren {-1}^{k + 1} } {q_k q_{k - 1} }$

Let $k$ be even.

Then:
 * $\exists s \in \N: k = 2 s$

and:
 * $C_{2 s} < C_{2 s - 1}$

Let $k$ be odd.

Then:
 * $\exists t \in \N: k = 2 t + 1$

and:
 * $C_{2 t + 1 } > C_{2 t}$

Now, consider any even convergent $C_{2 s}$ and any odd convergent $C_{2 t + 1}$, with $s, t \ge 0$.

If $2 s \le 2 t + 1$, then $2 s \le 2 t$, so:

If $2 s > 2 t + 1$, then $2 s - 1 \ge 2 t + 1$, so:

In either case:
 * $C_{2 s} < C_{2 t + 1}$

Hence the result.

Also see

 * Properties of Convergents of Continued Fractions