I'm working in some topics of Numerical Analysis and I founded a problem that sounds so interesting but I don't really know how to do it. The problem:
Let $a=(+.b_1b_2\dots)\times \beta^{p}$, $0<b_1<\beta$, $0\leq b_k<\beta$ for $k=2,3,\dots$ and $p\in\mathbb{Z}$. Take the sequence $\{q_n \}_{n\in\mathbb{N}}\subseteq\mathbb{Q}$ defined by $q_n=(+.b_1b_2\dots b_n)\times \beta^{p}$. Then
- Prove that $\{ q_n\}_{n\in\mathbb{N}}$ is a bounded and increasing sequence. Thus, is convergent to some $a^*\in\mathbb{R}$. In fact, $a^*=\sup\{q_n\mid n\in\mathbb{N}\}$
- Prove that $0\leq a-q_n\leq \beta^{p-n}$. Thus $q_n\to a$ and therefore $a=a^*$.
My attempt: first of all, I'm so confussed with the notation $(+.b_1b_2\dots)\times\beta^p$ because the general form of a floating number is $\pm d_1.d_2d_3\dots d_t\times\beta^{e}$. I'm asuming that $(+.b_1b_2\dots)\times\beta^p= (+0.b_1b_2\dots)\times\beta^p$. Am I correct? If it is correct, then, firts, I think that is easy to prove that $\{q_n \}$ is increasing and bounded by above because for all $n\in\mathbb{N}$ is clear that $0\leq q_n\leq a$ Why? As $a$ and $q_n$ have the same exponent, then we can operate as usually do. Then $$a-q_n=(+.0000\dots0b_{n+1}b_{n+1}\dots)\times \beta^p\geq 0$$Moreover, if we take $n\in\mathbb{N}$ then $$q_{n+1}-q_n=(+.b_1b_2\dots b_nb_{n+1})\times\beta^p-(+.b_1b_2\dots b_n)=(+.0000\dots0b_{n+1})\times\beta^p\geq0$$Therefore $q_{n+1}\geq q_n$ and thus $\{q_n \}$ is increasing. By the Least-upper-bound property, $\{q_n \}$ converges to its supremum. Is it correct?
To prove that $0\leq a-q_n\leq \beta^{p-n}$, I don't know how. I really appreciate any help you can provide me! Thanks!