Denominators of Simple Continued Fraction are Strictly Increasing

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

Let $$q_1, q_2, q_3, \ldots$$ be its denominators.

Then:
 * With the possible exception of $$q_1 = q_2$$, the sequence $$\left \langle {q_n}\right \rangle$$ is strictly increasing.
 * For any SCF, $$\forall k > 5: q_k > k$$.

Proof
By definition of simple continued fraction, all partial quotients of $$\left[{a_1, a_2, a_3, \ldots}\right]$$ are positive, with the possible exception of $$a_1$$.

So $$q_1 = 1, q_2 = a_2$$ and $$q_3 = a_3 a_2 + 1$$ all satisfy $$1 = q_1 \le q_2 < q_3$$.

Suppose $$q_k > q_{k-1} \ge 1$$ for some $$k \ge 3$$.

Then $$q_{k+1} = a_{k+1} q_k + q_{k-1} \ge q_k + q_{k-1} \ge q_{k} + 1 > q_k$$.

So, by induction, $$\left \langle {q_n}\right \rangle$$ is strictly increasing except when possibly $$q_1 = q_2 = 1$$.

Now, since $$q_3 \ge 2$$, from above, $$q_{k+1} \ge q_k + q_{k-1}$$ shows that from $$q_4$$ onwards, the $$q_k$$s increase in steps of at least $$2$$.

As $$q_4 \ge 3$$, it follows that $$q_5 \ge 5$$.

The result follows.