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I'm trying to show that the continued fraction $[1,1,\dots]$ converges. After that it is easy to determine the limit, so I'm interested in a proof of convergence specifically.

I don't think it's difficult to prove using Banach's fixed point theorem, but I'm looking for a more elementary proof, hopefully one that doesn't use more than basic results about infinite sequences.

Danny
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  • I often think it's easier to first find what the limit could be, and then prove convergence after that. Proving convergence is often made easier if you know what it's supposed to converge to. – Arthur Nov 06 '18 at 18:51
  • Might be relevant https://math.stackexchange.com/a/1871818/269624 – Yuriy S Nov 06 '18 at 19:08

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Continued fractions converge by their structural properties. If we have $$ \alpha=[a_0;a_1,a_2,a_3,\ldots]$$ with $a_0\in\mathbb{N}$ and $a_{>0}\in\mathbb{N}^+$, by denoting through $$ \frac{p_n}{q_n}=[a_0;a_1,a_2,\ldots,a_n] $$ the $n$-th convergent of $\alpha$ we have that $\frac{p_{n+2}}{q_{n+2}}$ always belongs to the interval whose endpoints are $\frac{p_n}{q_n}$ and $\frac{p_{n+1}}{q_{n+1}}$. Additionally $\left|\frac{p_n}{q_n}-\frac{p_{n+1}}{q_{n+1}}\right|$ exactly equals $\frac{1}{q_n q_{n+1}}$ and $\{q_n\}_{n\geq 1}$ is a rapidly (exponentially) increasing sequence. It follows that both $\left\{\frac{p_{2n}}{q_{2n}}\right\}_{n\geq 1}$ and $\left\{\frac{p_{2n+1}}{q_{2n+1}}\right\}_{n\geq 0}$ are monotonic, convergent, and convergent to the same limit, $\alpha$.

Jack D'Aurizio
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    Nice answer! Might be good to add the forward linear recurrence for $q_n$ to show why it increases exponentially. $$q_n=a_n q_{n-1}+q_{n-2}$$ – Yuriy S Nov 06 '18 at 19:31