Convergence of Infinite Continued Fractions

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Recall from The nth Convergent of an Infinite Continued Fraction page that if $\langle a_0; a_1, a_2, ... \rangle$ is an infinite simple continued fraction with $a_0 \in \mathbb{Z}$ and $a_n \in \mathbb{N}$ for $n \geq 1$ then the $n^{\mathrm{th}}$ convergent of this infinite simple continued fraction is defined to be:

(1)
\begin{align} \quad r_n = \langle a_0; a_1, a_2, ..., a_n \rangle \end{align}

We say that $\langle a_0; a_1, a_2, ... \rangle$ converges if $\displaystyle{\lim_{n \to \infty} r_n}$ exists. We also proved the following identities:

(2)

We will now prove that every infinite simple continued fraction of the form above converges.

 Theorem 1: Let $\langle a_0; a_1, a_2, ... \rangle$ be an infinite simple continued fraction where $a_0 \in \mathbb{Z}$ and $a_n \in \mathbb{N}$ for $n \geq 1$. Then: a) If $(r_n)$ is the sequence of $n^{\mathrm{th}}$ convergents then $r_0 < r_2 < r_4 < ... < r_5 < r_3 < r_1$. b) $\displaystyle{\lim_{n \to \infty} r_n}$ exists and $\displaystyle{r_{2j} < \lim_{n \to \infty} r_n < r_{2j-1}}$ for all $j$.
• Proof of a) For all $j \geq 0$ we have that:
(3)
\begin{align} \quad r_j - r_{j-2} = \frac{(-1)^ja_j}{k_jk_{j-2}} \end{align}
• If $j$ is even, then $(-1)^j = 1$, and since $a_j > 0$ and $k_j, k_{j-2} > 0$, we have that:
(4)
\begin{align} \quad r_j - r_{j-2} > 0 \end{align}
• Hence $r_0 < r_2 < r_4 < ...$. $(*)$
• If $j$ is odd, then $(-1)^{j} = -1$, and since $a_j >0$ and $k_j, k_{j-2} > 0$, we have that:
(5)
\begin{align} \quad r_j - r_{j-2} < 0 \end{align}
• Hence $... < r_5 < r_3 < r_1$. $(**)$
• Now for all $j \geq 0$ we have that:
(6)
\begin{align} \quad r_j - r_{j-1} = \frac{(-1){j-1}}{k_jk_{j-1}} \end{align}
• If $j$ is even, then $(-1)^{j-1} = -1$, and since $k_j, k_{j-1} > 0$ we have that $r_j - r_{j-1} < 0$. Hence $r_j < r_{j+1}$ for all even $j$. This combined with $(*)$ and $(**)$ above show that:
(7)
\begin{align} \quad r_0 < r_2 < r_4 < ... < r_5 < r_3 < r_1 \quad \blacksquare \end{align}
• Proof of b) Consider the subsequences $(r_{2n})$ and $(r_{2n-1})$ of $(r_n)$. Observe from $(a)$ that $(r_{2n})$ is an increasing sequence bounded above by $r_1$ and so $(r_{2n})$ converges to some $M \in \mathbb{R}$. Similarly, $(r_{2n-1})$ is a decreasing sequence bound below by $r_0$ and so $(r_{2n-1})$ also converges to some $N \in \mathbb{R}$. For all $j$ we have that $r_{2j} \leq M$ and $L \leq r_{2j-1}$, and:
(8)
\begin{align} \quad |L - M| \leq |r_{2j} - r{2j-1}| = \frac{|(-1)^{j-1}|}{k_{2j}k_{2j-1}} = \frac{1}{k_{2j}k_{2j-1}} \end{align}
• But the righthand side tends to $0$ as $j \to \infty$. Hence $L = M$. Hence $\displaystyle{\lim_{n \to \infty} r_n}$ exists and is such that for all $j$:
(9)
\begin{align} \quad r_{2j} < \lim_{n \to \infty} r_n < r_{2j-1} \quad \blacksquare \end{align}