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


 * $(1): \quad$ The odd convergents form a strictly increasing sequence:


 * $C_1 < C_3 < C_5 < \cdots$


 * $(2): \quad$ The even convergents form a strictly decreasing sequence:


 * $C_2 > C_4 > C_6 > \cdots$


 * $(3): \quad$ Every even convergent is greater than every odd convergent.

Proof
Let $p_1, p_2, p_3, \ldots$ and $q_1, q_2, q_3, \ldots$ be its numerators and denominators.

From Difference between Adjacent Convergents But One of Simple Continued Fraction:
 * $(1): \quad \forall k \ge 3: C_k - C_{k - 2} = \dfrac {\left({-1}\right)^{k - 1} a_k} {q_k q_{k - 2} }$

From Difference between Adjacent Convergents of Simple Continued Fraction:
 * $(2): \quad \forall k \ge 2: C_k - C_{k - 1} = \dfrac {\left({-1}\right)^k} {q_k q_{k - 1} }$

By definition of simple continued fraction, $\forall k \ge 2: a_k > 0$.

From Convergents of Simple Continued Fraction are Rationals in Canonical Form:
 * $\forall k \ge 1: q_k > 0$

Also from Convergents of Simple Continued Fraction are Rationals in Canonical Form:
 * $\forall k \ge 3: \dfrac {\left({-1}\right)^{k - 1} a_k} {q_k q_{k - 2} }$

has the same sign as $\left({-1}\right)^{k-1}$.

So it is positive when $k$ is odd and negative when $k$ is even.

Putting $k = 2 r + 1$ in $(1)$ gives:
 * $\forall r \ge 1: C_{2 r + 1} - C_{2 r - 1} > 0$

Putting $k = 2 r$ in $(1)$ gives:
 * $\forall r > 1: C_{2 r} - C_{2 r - 2} < 0$

Next, also from Convergents of Simple Continued Fraction are Rationals in Canonical Form, we have that:
 * $\forall k \ge 2: \dfrac {\left({-1}\right)^k} {q_k q_{k - 2} }$

has the same sign as $\left({-1}\right)^{k}$.

Putting $k = 2 r$ in $(2)$ gives:
 * $\forall r \ge 1: C_{2 r} - C_{2 r - 1} > 0$

and putting $k = 2 r + 1$ in $(2)$ gives:
 * $\forall r \ge 1: C_{2 r + 1} - C_{2 r} < 0$

So the even convergent $C_{2 r}$ is larger than each of the adjacent odd convergents $C_{2 r + 1}$ and $C_{2 r - 1}$.

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

If $s \ge t$ then $C_{2 s} > C_{2 s + 1} \ge C_{2 t + 1}$ (as odd convergents form a strictly increasing sequence).

If $s < t$ then $C_{2 s} > C_{2 t} > C_{2 t + 1}$ (as even convergents form a strictly decreasing sequence).

In each case $C_{2 s} > C_{2 t + 1}$ as required.