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

Theorem
For any SCF:
 * The odd convergents form a strictly increasing sequence:


 * $$C_1 < C_3 < C_5 < \cdots$$


 * The even convergents form a strictly decreasing sequence:


 * $$C_2 > C_4 > C_6 > \cdots$$


 * Every even convergent is greater than every odd convergent.

Proof
Let $$\left[{a_1, a_2, a_3, \ldots}\right]$$ be a continued fraction expansion.

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

From Properties of Convergents of Continued Fractions, we have that:
 * $$(1) \qquad C_k - C_{k-2} = \frac {\left({-1}\right)^{k-1} a_k} {q_k q_{k-2}}$$ for $$k \ge 3$$.
 * $$(2) \qquad C_k - C_{k-1} = \frac {\left({-1}\right)^k} {q_k q_{k-1}}$$ for $$k \ge 2$$.

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

From Properties of Convergents of Continued Fractions, we have that $$\forall k \ge 1: q_k > 0$$.

Then we note, also from Properties of Convergents of Continued Fractions, that:
 * $$\forall k \ge 3: \frac {\left({-1}\right)^{k-1} a_k} {q_k q_{k-2}}$$

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

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

Next, also from Properties of Convergents of Continued Fractions, we have that:
 * $$\forall k \ge 2: \frac {\left({-1}\right)^k} {q_k q_{k-2}}$$

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

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

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

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

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

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

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

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