Simple Infinite Continued Fraction Converges

Theorem

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

Then $C$ converges.

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

Let $\epsilon > 0$.

For $m > n \ge \max \set {5, \dfrac 1 \epsilon}$:

$\ds \size {C_m - C_n}$

$\le$

$\ds \size {C_m - C_{m - 1} } + \cdots + \size {C_{n + 1} - C_n}$

$\ds $

$=$

$\ds \frac 1 {q_m q_{m - 1} } + \cdots + \frac 1 {q_{n + 1} q_n}$

$\ds $

$<$

$\ds \frac 1 {m \paren {m - 1} } + \cdots + \frac 1 {\paren {n + 1} n}$

$\ds $

$=$

$\ds \sum_{k \mathop = n}^{m - 1} \frac 1 {k \paren {k + 1} }$

$\ds $

$=$

$\ds \sum_{k \mathop = n}^{m - 1} \paren {\frac 1 k - \frac 1 {k + 1} }$

$\ds $

$=$

$\ds \frac 1 n - \frac 1 m$

$\ds $

$<$

$\ds \frac 1 n$

$\ds $

$\le$

$\ds \epsilon$

So $\sequence {C_n}$ is indeed a Cauchy sequence.

Hence the result.

$\blacksquare$