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 § 10, but I just can't seem to figure this one out.
Exercise 10.3
For a decimal expansion K.d1d2d3d4⋯, let (sn) be defined as in Discussion 10.3. Prove sn<K+1 for all n∈N. Hint: 910+9102+⋯+910n=1−110n.
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.d1d2d3d4⋯, where K is a nonnegative integer and each dj belongs to {0,1,2,3,4,5,6,7,8,9}. Let
sn=K+d110+d2102+⋯+dn10n.
Then (sn) is an increasing sequence of real numbers, and (sn) is bounded [by K+1, in fact]. So by Theorem 10.2, (sn) converges to a real number we traditionally write as K.d1d2d3d4⋯. For example, 3.3333⋯ represents
limn→∞(3+310+3102 +⋯+310n).
To calculate this limit, we borrow the following fact about geometric series:
limn→∞a(1+r+r2+⋯+rn)=a1−rfor|r|<1.
In our case, a=3 and r=110, so 3.3333⋯ represents 31−110=103, as expected. Similarly, 0.9999⋯ represents
limn→∞(910+9102+⋯+910n)=9101−110=1.
Thus 0.9999⋯ and 1.0000⋯ 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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