# Integral part of an infinite simple continued fraction

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.

• Just write out a few continued fractions, maybe a few convergents each for $\pi,$ then $e,$ then Euler's $\gamma.$ Things become clear that are obscured by the copious symbols used Jan 7, 2021 at 1:55
• I know it is pretty obvious but proving it... not so much for me. Jan 7, 2021 at 2:00
• There is a space between obvious and proof in which we try to figure out what we ought to be attempting to prove. Continued fractions have too much machinery for most beginners to appreciate everything in symbols. Write out a few convergents in a few examples, and you will see enough to find something fairly specific that could be proved. I suppose I should direct your attention to where the convergents lie relative to the constant, above or below Jan 7, 2021 at 2:04
• I know even convergents are below and strictly increase to the infinite continued fraction and odd convergents are above and strictly decrease to the infinite continued fraction. So I suppose I could argue that we reach a point where the convergents all have the same integral part (this is possible since infinite continued fractions are irrational numbers). Thus the limit of the sequence of convergents must clearly have the same integral part and those of the 'big' convergents. Jan 7, 2021 at 2:14
• good. Please do the beginnings of a few cf's for positive numbers... just the convergents coming from $a_0$ then $a_1$ then $a_2$ for, say, $\pi$ and $e.$ Every subject in mathematics that can be taught at all has sample calculations that give a concrete feel. Here, it is actual convergents, writing down some with specific numbers, getting decimal versions of the convergents and seeing how they compare to the eventual limit. Not from a remembered theorem, rather from new experience Jan 7, 2021 at 2:22

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 $$a_0 = \lfloor{\alpha}\rfloor$$.