July 23, 2021

Continued Fractions, Attempt 3, Part 8: A Real Case of Infinite Continued Fractions

1. 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 any $n\in\N$, let $r_n$ be a non-negative real number and define $s_n\in\R$ by \[ s_n=s_0+\sum_{k=0}^{n-1}(-1)^kr_k \] where $s_0$ is an arbitrary constant.

Lemma 8.1. If $r_n$ decreases monotonically as $n\to\infty$, that is, $r_n \ge r_{n+1}$ for any $n\in\N$, then $s_{2n}$ increases monotonically and $s_{2n+1}$ decreases monotonically. Moreover, for any $(j,k)\in\N^2$, we have $s_{2j}\le s_{2k+1}$.

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 8.2. $s_{2n}$ 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 \[ \lim_{n\to\infty}s_{2n}=\sup\{s_{2n}\,|\,n\in\N\}. \] Similarly, \[ \lim_{n\to\infty}s_{2n+1}=\inf\{s_{2n+1}\,|\,n\in\N\}. \] By definition, for any $k\in\N$, we have \[ \sup\{s_{2n}\,|\,n\in\N\}\le s_{2k+1} \] and therefore \[ \sup\{s_{2n}\,|\,n\in\N\}\le\inf\{s_{2n+1}\,|\,n\in\N\}. \]

Theorem 8.3. 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 any $\epsilon>0$ there exist $n_0,n_1\in\N$ such that for any $j>n_0$ holds $|s_{2j}-\sigma|<\epsilon$ and for any $k>n_1$ holds $|s_{2k+1}-\sigma|<\epsilon$. This means for any $n>2\max(n_0,n_1)+1$ holds $|s_n-\sigma|<\epsilon$ and therefore $\lim_{n\to\infty}s_n=\sigma$.

2. A Real Case of Infinite Continued Fractions

Let $a_0$ be a real number, $a_1$ a positive real number, and $a_{n+2}$ a non-negative real number for any $n\in\N$. Let $\gamma$ be a simple infinite continued fraction $((a_0,1),(a_1,1),(a_2,1),\ldots)$ and $c_n$ the $n$-th convergent of $\gamma$.

Lemma 8.4. For any $n\in\N$, define $d_n\in\whC$ by \[ d_n=(-1)^n(c_{n+2}-c_{n+1}). \] Then, $d_n$ is actually a positive real number and decreases monotonically as $n\to\infty$.

Proof. Let us write $c_n=p_n/q_n$ as in Part 6. For any $n\in\N$, we have \[ \phantom{\text{(8.1)}}\hspace{6em} q_{n+2}=a_{n+1}q_{n+1}+q_n, \quad q_1=1, \quad q_0=0, \hspace{6em}\text{(8.1)} \] and therefore $q_{2n+4}\ge q_{2n+2}>q_0=0$ and $q_{2n+3} \ge q_{2n+1} \ge q_1=1$. By Corollary 6.5 in Part 6, we have \[ d_n=\frac{1}{q_{n+2}q_{n+1}} \] and therefore $d_n$ is a positive real number. Also, we have \[ \frac{d_{n+1}}{d_n}=\frac{q_{n+1}}{q_{n+3}}\le1 \] and therefore $d_n$ decreases monotonically as $n\to\infty$.

Corollary 8.5. For any $n\in\N$, define $c'_n\in\R$ by \[ c'_n=a_0+\sum_{k=0}^{n-1}(-1)^kd_k, \quad c'_0=a_0. \] Of course, $c'_n=c_{n+1}$. Then, $(c'_{2n})_{n\in\N}$ is monotonically increasing and $(c'_{2n+1})_{n\in\N}$ monotonically decreasing, as $n\to\infty$. Moreover, $c'_{2j}\le c'_{2k+1}$ for any $(j,k)\in\N^2$.

Proof. By Lemma 8.1.

Theorem 8.6. $c'_n$ converges to a finite real value if and only if \[ \sum_{k=0}^{\infty} a_{k+1} \] converges to $\infty$.

Proof. Assume $\sum_{k=0}^{\infty}a_{k+1}$ converges to $\infty$. By $\text{(8.1)}$, 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 $n\to\infty$, $d_n$ converges to $0$ and by Theorem 8.3 $c'_n$ converges to a finite real value.

Conversely, assume $\sum_{k=0}^{\infty}a_{k+1}$ does not converge to $\infty$. This implies it converges to a finite value and $\prod_{k=0}^{\infty}(1+a_{k+1})$ converges to a finite value. By $\text{(8.1)}$, we have \[ \begin{aligned} q_{n+2}+q_{n+1}&=(1+a_{n+1})q_{n+1}+q_n \\ &\le (1+a_{n+1})(q_{n+1}+q_n) \\ &\le (q_1+q_0)\prod_{k=0}^{n}(1+a_{k+1}) \\ &=\prod_{k=0}^{\infty}(1+a_{k+1}). \end{aligned} \] Therefore, letting $P=\prod_{k=0}^{\infty}(1+a_{k+1})$, we have \[ d_n=\frac{1}{q_{n+2}q_{n+1}}\ge\left(\frac{2}{q_{n+2}+q_{n+1}}\right)^2\ge\frac{4}{P^2}. \] This means $c'_n$ does not converge in $\whC$.


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