# Finding the limit of a sequence (different decimal expansions for the same real number)

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 $$d_n = a_0 \bullet a_1a_2...a_n$$.

• is the equation in the first line meant to be $a_{n+1} = a_n + (\frac{1}{2})^n$? – videlity May 4 '17 at 6:58
• @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. – jenkat May 4 '17 at 12:17
• 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 $a_n = 3-(\frac{1}{2})^{n-1}$. – videlity May 5 '17 at 3:11

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.