July 14, 2020

Continued Fractions, Part 4: Real Convergence

This is the fourth article in the continued fractions series, following Part 3.

Alternating Series

Let us recall the well-known condition for convergence of alternating series. In fact, the condition is two-fold and handling them separately is useful for continued fractions.

For all $n\in\N$, let $r_n$ be non-negative real numbers and define $s_n$ by \[ s_n=\sum_{k=0}^{n-1}(-1)^kr_k, \quad s_0=0. \]

Lemma 4.1. If $r_n$ decreases monotonically as $n\to\infty$, that is, $r_n \ge r_{n+1}$ for all $n\in\N$, then $s_{2n}$ increases monotonically and $s_{2n+1}$ decreases monotonically. Moreover, $s_{2j}\le s_{2k+1}$ for all $(j,k)\in\N^2$.

Proof. We have \[ s_{2n+2}-s_{2n}=-r_{2n+1}+r_{2n}\ge0 \] and \[ s_{2n+3}-s_{2n+1}=r_{2n+2}-r_{2n+1}\le0. \] If $2j<2k+1$, then \[ s_{2j} \le s_{2k} \le s_{2k}+r_{2k}=s_{2k+1}. \] If $2j>2k+1$, then \[ s_{2j}=s_{2j-1}-r_{2j-1}\le s_{2j-1} \le s_{2k+1}. \]

Corollary 4.2. $s_{2n}$ and $s_{2n+1}$ converge to finite values as $n\to\infty$, and $\lim_{n\to\infty}s_{2n}\le\lim_{n\to\infty}s_{2n+1}$.

Proof. $s_{2n}$ increases monotonically and bounded from above by, say, $s_1$. Hence \[ \lim_{n\to\infty}s_{2n}=\sup\{s_{2n}\}. \] Similarly, \[ \lim_{n\to\infty}s_{2n+1}=\inf\{s_{2n+1}\}. \] By definition, we have $\sup\{s_{2n}\}\le s_{2k+1}$ for all $k\in\N$ and therefore $\sup\{s_{2n}\}\le\inf\{s_{2n+1}\}$.

Theorem 4.3. If $r_n$ additionally satisfies the condition $\lim_{n\to\infty}r_n=0$, then \[ \lim_{n\to\infty}s_n=\lim_{n\to\infty}s_{2n}=\lim_{n\to\infty}s_{2n+1}. \]

Proof. We have \[ \lim_{n\to\infty}(s_{2n+1}-s_{2n})=\lim_{n\to\infty}r_{2n}=0 \] and therefore $\lim_{n\to\infty}s_{2n}=\lim_{n\to\infty}s_{2n+1}$. Let $\sigma$ denote this limit. For all $\epsilon>0$ there exist $n_0,n_1\in\N$ such that for all $j>n_0$ holds $|s_{2j}-\sigma|<\epsilon$ and for all $k>n_1$ holds $|s_{2k+1}-\sigma|<\epsilon$. This means for all $n>2\max(n_0,n_1)+1$ holds $|s_n-\sigma|<\epsilon$ and therefore $\lim_{n\to\infty}s_n=\sigma$.

Real Convergence

Let $a_0$ be a real number, $a_1$ a positive real number, and $a_{n+2}$ a non-negative real number for all $n\in\N$. Let $\kappa$ be $[a_0,a_1,a_2,\ldots]$ and $\{c_n\}_{n\in\N}$ denote convergents of $\kappa$.

Lemma 4.4. For all $n\in\N$, $(-1)^n(c_{n+2}-c_{n+1})$ is a positive real number and decreases monotonically as $n\to\infty$.

Proof. By $\text{(1.1b)}$ in Part 1, we have $q_{2n+4}\ge q_{2n+2}>q_0=0$ and $q_{2n+3} \ge q_{2n+1} \ge q_1=1$ for all $n\in\N$. This implies $q_{n+1}\ge\min(q_2,q_1)>0$ and thus $c_{n+1}\ne\infty$ for all $n\in\N$. By the corollary in Part 1, we have \[ (-1)^n(c_{n+2}-c_{n+1})=\frac{1}{q_{n+2}q_{n+1}}>0. \] and \[ \frac{(-1)^{n+1}(c_{n+3}-c_{n+2})}{(-1)^n(c_{n+2}-c_{n+1})}=\frac{q_{n+1}}{q_{n+3}}\le1. \] for all $n\in\N$.

Corollary 4.5. $\{c_{2n+1}\}_{n\in\N}$ is monotonically increasing and $\{c_{2n+2}\}_{n\in\N}$ monotonically decreasing, as $n\to\infty$. Moreover, $c_{2j+1}\le c_{2k+2}$ for all $(j,k)\in\N^2$.

Theorem 4.6. $c_n$ converges to a finite real value if and only if \[ \sum_{n=0}^{\infty} a_n \] converges to $\infty$.

Proof. Assume $\sum_{n=0}^{\infty} a_n$ converges to $\infty$. By $\text{(1.1b)}$ in Part 1, we have \[ \begin{aligned} q_{n+2}q_{n+1}&=q_{n+1}q_n+a_{n+1}q_{n+1}^2 \\ &=q_1q_0+\sum_{k=0}^na_{k+1}q_{k+1}^2 \\ &\ge\min(q_2,q_1)^2\sum_{k=0}^na_{k+1}. \end{aligned} \] Therefore, as $n\to\infty$, $q_{n+2}q_{n+1}$ converges to $\infty$ and by Theorem 4.3 $c_n$ converges to a finite real value.

Conversely, assume $\sum_{n=0}^{\infty} a_n$ does not converge to $\infty$. This implies it converges to a finite value and $\prod_{n=0}^{\infty}(1+a_n)$ converges to a finite value. By $\text{(1.1b)}$ in Part 1, we have \[ \begin{aligned} q_{n+2}+q_{n+1}&=(1+a_{n+1})q_{n+1}+q_n \\ &\le (1+a_{n+1})q_{n+1}+(1+a_{n+1})q_n \\ &=(1+a_{n+1})(q_{n+1}+q_n) \\ &\le (q_1+q_0)\prod_{k=0}^{n}(1+a_{k+1}) \\ &\le \prod_{k=0}^{\infty}(1+a_{k+1}). \end{aligned} \] Therefore, letting $S=\prod_{k=0}^{\infty}(1+a_{k+1})$, for all $n\in\N$ we have $q_{n+1}\le S$ and \[ (-1)^n(c_{n+2}-c_{n+1})=\frac{1}{q_{n+2}q_{n+1}}\ge\frac{1}{S^2}>0. \] This means $c_n$ does not converge in $\whC$.


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