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

Theorem
Let $n \in \N \cup \{\infty\}$ be an extended natural number.

Let $(a_0, a_1, \ldots)$ 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 $(C_0, C_1, \ldots)$ be its sequence of convergents.

Every even convergent is strictly smaller than every odd convergent.

Proof
Let $k \geq 1$.

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

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

Thus if $k=2s$ is even, $C_{2s} < C_{2s-1}$.

And if $k=2t+1$ is odd, $C_{2t+1} > C_{2t}$.

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

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

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

In any case, $C_{2 s} < C_{2 t + 1}$ as required.

Also see

 * Properties of Convergents of Continued Fractions