Sequence of Best Rational Approximations to Square Root of 2

Theorem
A sequence of best rational approximations to the square root of $2$ starts:
 * $\dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$

where:
 * the numerators are half of the Pell-Lucas numbers, $\frac 12 Q_n$
 * the denominators are the Pell numbers $P_n$

starting from $\dfrac {\tfrac12 Q_1} {P_1}$.

Proof
Let $(a_0, a_1, \ldots)$ be the continued fraction expansion of $\sqrt 2$.

By Continued Fraction Expansion of Root 2:
 * $\sqrt 2 = \left[{1, \left \langle{2}\right \rangle}\right] = \left[{1, 2, 2, 2, \ldots}\right]$

From Convergents are Best Approximations, the convergents of $\left[{1, \left \langle{2}\right \rangle}\right]$ are the best rational approximations of $\sqrt 2$.

Let $(p_n)_{n \geq 0}$ and $(q_n)_{n \geq 0}$ be the numerators and denominators of the continued fraction expansion of $\sqrt 2$.

Then $\dfrac {p_n} {q_n}$ is the $n$th convergent of $\left[{1, \left \langle{2}\right \rangle}\right]$.

By Convergents of Simple Continued Fraction are Rationals in Canonical Form, $p_n$ and $q_n$ are coprime for all $n\geq 0$.

It remains to show that for all $n \geq 1$:
 * $Q_n = 2p_{n-1}$
 * $P_n = q_{n-1}$

It suffices to prove that they satisfy the same recurrence relation.

By definition:

so that $(Q_1, P_1) = (2, 1) = (2p_0, q_0)$.

so that $(Q_2, P_2) = (6, 2) = (2p_1, q_1)$.

The result follows by definition of Pell numbers and Pell-Lucas numbers.