Every infinite regular continued fraction converges to a unique real number

Since every partial quotient after the first is at least $1$. [given, F1, induction, algebra] The denominators satisfy $q_{n+1}=a_{n+1}{q}_n+q_{n-1}\ge q_n+q_{n-1}>q_n>0$ for $n\ge 1$, with $q_0=1$ and $q_1=a_1\ge 1$. Thus $(q_n)$ is nondecreasing from $q_0$ onward and strictly increasing from $q_1$ onward, and induction on $n$ gives $q_n\ge n$ for $n\ge 1$.

By [F1], the signs of $c_n-c_{n-1}$ alternate, and by step 1.1 their absolute values strictly decrease. [F1, step 1.1, algebra] Hence $c_{2m}<c_{2m+1},c_{2m+2}=c_{2m+1}-(c_{2m+1}-c_{2m+2})>c_{2m},$ and similarly $c_{2m+3}=c_{2m+2}-(c_{2m+2}-c_{2m+3})<c_{2m+1}\mathrm{.}$ So the even convergents increase, the odd convergents decrease, and every even convergent is below every odd convergent.

Put $d_n:=1\mathrm{/}(q_n{q}_{n+1})$. [F2, step 1.1, algebra] Step 1.1 gives $d_n\le 1\mathrm{/}(n(n+1))$ for $n\ge 1$, and the Archimedean property [F2] therefore implies $d_n\to 0$. For $\epsilon >0$ choose $N\ge 1$ with $1\mathrm{/}N<\epsilon$, then for $n\ge N$, $0<d_n\le \frac{1}{n(n+1)}\le \frac{1}{n}\le \frac{1}{N}<\epsilon \mathrm{.}$

Let $E:=\{c_{2m}:m\ge 0\}$. [step 2.1, given] By step 2.1 the set $E$ is nonempty and bounded above by $c_1$, so [F3] gives a real number $x:=\sup E$.

If $n=2m+1$ is odd, then $c_{n-1}\in E$ and step 2.1 gives. [step 2.1, step 2.2, F1] $c_{n-1}\le x\le c_n$, so $0\le c_n-x\le c_n-c_{n-1}=d_{n-1}\mathrm{.}$ If $n=2m$ is even, then $c_n\in E$ and step 2.1 gives $c_n\le x\le c_{n+1}$, so $0\le x-c_n\le c_{n+1}-c_n=d_n\mathrm{.}$ Since the right-hand sides tend to $0$ by step 2.2, both subsequences converge to $x$.

If $y$ were another real with both subsequences converging to $y$, then for every $m$. [step 3.2, algebra] $\mathrm{\mid }x-y\mathrm{\mid }\le \mathrm{\mid }x-c_{2m}\mathrm{\mid }+\mathrm{\mid }{c}_{2m}-y\mathrm{\mid },$ and the right-hand side tends to $0$ as $m\to \mathrm{\infty }$ by step 3.2. Hence $x=y$, so the value is unique.