Let $[a_0,a_1,a_2,\ldots]$ be an infinite simple continued fraction.
How can one show that $a_0$ is the integral part of $[a_0,a_1,a_2,\ldots]$?
There is a similar theorem for finite continued fractions with length $n\geq 1$ and $a_n \geq 2$ that I know, however I don't know how to extend this to the infinite case. The infinite continued fraction is the limit of the sequence of its convergents but as each convergent will not always have $2$ as its last partial coefficient, I don't know how to proceed.
For $n \geq 2$, every finite continued fraction $[a_2,a_3,\ldots,a_n]$ is at least $a_2$. Therefore the infinite continued fraction $\beta := [a_2,a_3,\ldots]$, which is defined to the the limit of the finite continued fractions $[a_2,a_3,\ldots,a_n]$ is at least $a_2$: $\beta \geq a_2 > 0$.
Setting $\alpha = [a_0,a_1,a_2,\ldots]$, we have $$ \alpha = a_0 + \frac{1}{a_1 + 1/\beta}. $$ Since $\beta > 0$ we have $0 < 1/\beta$, so $0 < a_1 < a_1 + 1/\beta$, so $$ 0 < \frac{1}{a_1 + 1/\beta} < \frac{1}{a_1} \leq 1 \Longrightarrow 0 < \frac{1}{a_1 + 1/\beta} < 1. $$ Add $a_0$ to each term in the last inequality and you get $a_0 < \alpha < a_0 + 1$, so $a_0 = \lfloor{\alpha}\rfloor$.