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From the book Elementary Analysis: The Theory of Calculus by Kenneth A. Ross


I've been working on the exercises at the end of Chapter 2 $\S$ 10, but I just can't seem to figure this one out.

Exercise 10.3

For a decimal expansion $K.d_1d_2d_3d_4\cdots$, let $(s_n)$ be defined as in Discussion 10.3. Prove $s_n < K + 1$ for all $n \in \mathbb N$. Hint: $\frac{9}{10} + \frac{9}{10^2} + \cdots + \frac{9}{10^n} = 1 - \frac{1}{10^n}$.

Here is Discussion 10.3:

10.3 Discussion of Decimals

We restrict our attention to nonnegative decimal expansions and nonnegative real numbers. From our point of view, every nonnegative decimal expansion is shorthand for the limit of a bounded increasing sequence of real numbers. Suppose we are given a decimal expansion $K.d_1d_2d_3d_4\cdots$, where $K$ is a nonnegative integer and each $d_j$ belongs to $\{0,1,2,3,4,5,6,7,8,9\}$. Let

$$s_n = K + \frac{d_1}{10} + \frac{d_2}{10^2} + \cdots + \frac{d_n}{10^n}\tag{1}\label{1}.$$

Then $(s_n)$ is an increasing sequence of real numbers, and $(s_n)$ is bounded [by $K + 1$, in fact]. So by Theorem 10.2, $(s_n)$ converges to a real number we traditionally write as $K.d_1d_2d_3d_4\cdots$. For example, $3.3333\cdots$ represents

$$\lim_{n\to\infty} \left(3 + \frac{3}{10} + \frac {3}{10^2}\ + \cdots + \frac{3}{10^n}\right).$$

To calculate this limit, we borrow the following fact about geometric series:

$$\lim_{n\to\infty} a\left(1 + r + r^2 + \cdots + r^n\right) = \frac{a}{1-r} \quad \text{for} \quad |r| < 1 \tag{2}\label{2}.$$

In our case, $a = 3$ and $r = \frac{1}{10}$, so $3.3333\cdots$ represents $\frac{3}{1-\frac{1}{10}}=\frac{10}{3}$, as expected. Similarly, $0.9999\cdots$ represents

$$\lim_{n\to\infty} \left(\frac{9}{10} + \frac{9}{10^2} + \cdots + \frac{9}{10^n}\right) = \frac{\frac{9}{10}}{1-\frac{1}{10}} = 1.$$

Thus $0.9999\cdots$ and $1.0000\cdots$ are different decimal expansions that represent the same number.

This section in the book proved two important theorems (and the related lemmas/corollaries): (1) All bounded monotone sequences converge; and (2) A sequence is a convergent sequence if and only if it is a Cauchy sequence. So I'm wondering how I would use those theorems in this proof? Or would I? Would I have to use the ideas of lim sup and lim inf? Geometric series?

Thank you for your time.

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$s_{n} < K +1 \leftrightarrow K + d_{1}/10 + d_{2}/10 + ... d_{n}/10 < K +1 \leftrightarrow d_{1}/10 + d_{2}/10 + ... d_{n}/10 < 1 \leftarrow (by \quad hint) 1- 1/10^{n} < 1 \leftrightarrow 0 < 1/10^{n} $ with the last statement being trivially true. We can use the hint as taking all the $d_{j}$ values to be 9 is the biggest $s_{n}$ can be.

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  • $\begingroup$ So the proof doesn't require any use of the aforementioned theorems. I guess I was just overthinking the problem. Thanks. $\endgroup$ – Zach Franssen Apr 12 '17 at 7:08
  • $\begingroup$ unfortunately in this case the hint seems to do all the "analysis" for you, the exercise would have been better without the hint really. $\endgroup$ – Hugh Kinnear Apr 13 '17 at 13:27

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