Simple Infinite Continued Fraction Converges

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Theorem

Let $C = (a_0, a_1, \ldots)$ be a simple infinite continued fraction in $\R$.


Then $C$ converges.


Proof

We need to show that for any SICF its sequence of convergents $\left \langle {C_n}\right \rangle$ always tends to a limit.


Several techniques can be used here, but a quick and easy one is to show that $\left \langle {C_n}\right \rangle$ is a Cauchy sequence.

By Difference between Adjacent Convergents of Simple Continued Fraction:

$\left|{C_{k + 1} - C_k}\right| = \dfrac 1 {q_{k + 1} q_k}$


From Lower Bounds for Denominators of Simple Continued Fraction:

$q_k > k$

for sufficiently large $k$, $q_k > k$.

So:

$\dfrac 1 {q_{k + 1} q_k} < \dfrac 1 {\left({k + 1}\right) k} < \frac 1 {k^2}$

But $\left \langle {\dfrac 1 {k^2} }\right \rangle$ is a basic null sequence.

So by the Squeeze Theorem:

$\dfrac 1 {q_{k + 1} q_k} \to 0$

as $k \to \infty$.

So $\left \langle {C_n}\right \rangle$ is indeed a Cauchy sequence.


Hence the result.

$\blacksquare$