Exercise 7.4.4 ($ \lim_{n o \infty} \langle a_0,a_1,\ldots,a_n,b_1,b_2,b_3,\ldots angle = \xi$)

Question

Let $\xi$ be an irrational number with continued fraction expansion $\langle a_0,a_1,a_2,a_3,\ldots angle$. Let $b_1,b_2,b_3,\ldots$ be any finite or infinite sequence of positive integers. Prove that
$$\lim_{n	o \infty} \langle a_0,a_1,a_2,\ldots,a_n, b_1,b_2,b_3,\ldots angle = \xi.$$

Richard Ganaye · Accepted Answer

We write $h_i/k_i$ the convergents of $\xi$. By Theorem 7.10,
\begin{align}
\lim_{n	o \infty} \left( \xi -  \frac{h_n}{k_n}ight) = 0.
\end{align}
Put $\xi_n = \langle a_0,a_1,\ldots,a_n,b_1,b_2,b_3,\ldotsangle$ and $\eta = \langle b_1, b_2,b_3,\ldots angle>0$. then $\xi_n$ has the same convergents as $\xi$ up to $h_n/k_n$, and by Theorem 7.10
$$\xi_{n} =   \langle a_0,a_1,\ldots,a_n,\eta angle.$$
Then, as in (7.9), for all $n \geq 0$,
\begin{align*}
\xi_n - \frac{h_n}{k_n} &= \frac{\eta h_{n} + h_{n-1}}{ \eta k_{n} + k_{n-1}} - \frac{h_n}{k_n}\
&= \frac{-(h_n k_{n-1} - h_{n-1}k_n)}{k_n(\eta k_{n} + k_{n-1})}\
&= \frac{(-1)^n}{k_n(\eta k_{n} + k_{n-1})}\
\end{align*}
Since $\eta>0$,
\begin{align*}
\left | \xi_n - \frac{h_n}{k_n} ight | &= \frac{1}{k_n(\eta k_{n} + k_{n-1})}\
&< \frac{1}{k_n k_{n-1}}.
\end{align*}
The sequence of integers $(k_n)_{n\in \N}$ is strictly increasing, thus $\lim_{n	o \infty} \frac{1}{k_n k_{n-1}} = 0$, therefore
\begin{align}
\lim_{n 	o \infty} \left( \xi_n - \frac{h_n}{k_n}ight)= 0.
\end{align}
Since 
$$\xi - \xi_n = \left( \xi -  \frac{h_n}{k_n}ight) -  \left( \xi_n - \frac{h_n}{k_n}ight),$$
the limits (1) and (2) show that
$$\lim_{n 	o \infty} (\xi - \xi_n) = 0,$$
so
$$\xi = \lim_{n	o \infty}  \langle a_0,a_1,\ldots,a_n,b_1,b_2,b_3,\ldotsangle.$$
\end{proof}

Exercise 7.4.4 ($ \lim_{n\to \infty} \langle a_0,a_1,\ldots,a_n,b_1,b_2,b_3,\ldots\rangle = \xi$)

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Exercise 7.4.4 ($ \lim_{n\to \infty} \langle a_0,a_1,\ldots,a_n,b_1,b_2,b_3,\ldots\rangle = \xi$)

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