Relative Sizes of Convergents of Simple Continued Fraction

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:

• $(1) \qquad C_k - C_{k-2} = \dfrac {\left({-1}\right)^{k-1} a_k} {q_k q_{k-2}}$ for $k \ge 3$.
• $(2) \qquad C_k - C_{k-1} = \dfrac {\left({-1}\right)^k} {q_k q_{k-1}}$ 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:

$\forall k \ge 3: \dfrac {\left({-1}\right)^{k-1} a_k} {q_k q_{k-2}}$

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 $k = 2r + 1$ 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:

$\forall k \ge 2: \dfrac {\left({-1}\right)^k} {q_k q_{k-2}}$

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.

$\blacksquare$