In the following proof that $\sqrt2$ is irrational, I cannot make sense of why $\frac{q a_n + p b_n}{q} \geqslant \frac{1}{q}$.
The sums corresponding to $a_n$ and $b_n$ are alternating sums, so why must $q a_n + p b_n \geqslant 1$ ?
Proof:
Suppose $\sqrt2 = \frac{p}{q} \in \mathbb{Q}$. On the one hand, $0 < (\sqrt{2} - 1)^{n} < 1$. Thus,
$$ \lim_{n \to \infty} (\sqrt{2} - 1)^{n} = 0 $$
On the other hand, the binomial expansion of $(\sqrt{2} - 1)^{n}$ is given by
$$ (\sqrt{2} - 1)^{n} = \sum_{k = 0}^{n} {n \choose k}(\sqrt{2})^{k}(-1)^{n-k} $$ $$ = \sum_{\substack{k = 0 \\ k \text{ even}}}^{n} {n \choose k}(2)^{\frac k2}(-1)^{n-k} + \sqrt2 \sum_{\substack{k = 0 \\ k \text{ odd}}}^{n} {n \choose k}(2)^{\frac{k-1}{2}}(-1)^{n-k} $$ $$ = a_n + b_n \sqrt2 $$
Finally, we see that
$$ (\sqrt{2} - 1)^{n} = a_n + b_n \sqrt2 = a_n + b_n \frac{p}{q} = \frac{q a_n + p b_n}{q} \geqslant \frac{1}{q} $$
Contradicting the fact that $(\sqrt{2} - 1)^{n}$ tends to $0$ as $n \to \infty$
QED