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I am working through Foundations of Mathematics (2015) by Ian Stewart and David Tall. On page 39, the authors provide the following example. This is a distinct question from my earlier question regarding the same page. Note that decimal points are indicated by $\bullet$ to distinguish them from ...

Example 2.13: Suppose $a_1 = 1$ and in general $a_{n+1} = a_n + (\frac{1}{2})^{n-1}$, then trivially $(a_n)$ is increasing and a calculation gives $a_n = 2 - (\frac{1}{2})^{n-1}$, so the sequence is bounded above by 2. Using the same method to calculate the decimal expansion using definition (2.2)... the limit of the sequence $(a_n)$ is then found to be:

$b_0 \bullet b_1b_2...b_n... = 1 \bullet 99...9...$

Where definition 2.2 is a way to find the decimal expansion of a real number:

$a_0 \bullet a_1a_2...a_n < x \le a_0 \bullet a_1a_2...a_n + (\frac{1}{10})^n$

My understanding is that if $a_{n+1} = a_n + (\frac{1}{2})^{n-1}$, then $a_n = a_{n+1} - (\frac{1}{2})^{n-1}$. In the first blockquote, when the authors say "a calculation gives $a_n = 2 - (\frac{1}{2})^{n-1}$", do we assume this as a given? Otherwise, surely $a_{n+1}$ cannot equal $2$ for all cases?

A separate question is why, in blockquote 2, the limit is denoted using $b_n$ notation when the original sequence is denoted by $a_n$ notation. Is this standard?

The authors provide the following example after the one above. I am including it in case it is relevant to Example 2.13:

To cover all cases, we introduce the following:

Definition 2.14: The value of an infinite decimal $a_0 \bullet a_1a_2...a_n...$ is the limit $l$ of the sequence $(d_n)$ of decimals to $n$ decimal places, where $d_n = a_0 \bullet a_1a_2...a_n$.

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  • $\begingroup$ is the equation in the first line meant to be $a_{n+1} = a_n + (\frac{1}{2})^n$? $\endgroup$ – videlity May 4 '17 at 6:58
  • $\begingroup$ @videlity: No, I checked the book and the equation is $a_{n+1} = a_n + (\frac{1}{2})^{n-1}$. I have found typos in the book before, so if the example makes sense with $a_{n+1} = a_n + (\frac{1}{2})^n$, then I'd appreciate it if you could explain. $\endgroup$ – jenkat May 4 '17 at 12:17
  • $\begingroup$ I don't think it makes sense with that. Since using that formula, $a_2 = a_1+1 = 2$ and $a_3 = a_2 + \frac{1}{2} = 2.5$ and $a_4 = 2.75$. So either the first equation should be what I wrote above, or $a_n = 3-(\frac{1}{2})^{n-1}$. $\endgroup$ – videlity May 5 '17 at 3:11
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The calculation isn't too difficult which is why they probably left it out. Note that $$a_1=1, \\a_2 = a_1+\frac{1}{2} = 1+\frac{1}{2}, \\a_3 = a_2 + \frac{1}{2^2} = 1+\frac{1}{2}+\frac{1}{2^2}$$ and in general, $$a_n = 1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-1}}. $$ This is an geometric progression with common factor $r=\frac{1}{2}$ and using the formula, we get, $$a_n = \frac{1-\frac{1}{2^n}}{1-\frac{1}{2}} = 2-\left(\frac{1}{2}\right)^{n-1}.$$

This answers your first question. I'm not sure about the other question. I think why they used $b_i$ is just because they used $a_i$ earlier and they stand for different things. $b_i$ stands for the $i$th digit in the decimal, while $a_i$ is a term in the sequence.

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