Thursday, 8 January 2015

real analysis - For a decimal expansion K.d1d2d3d4cdots, prove sn<K+1 for all ninmathbbN

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 nN. Hint: 910+9102++910n=1110n.




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:



limna(1+r+r2++rn)=a1rfor|r|<1.




In our case, a=3 and r=110, so 3.3333 represents 31110=103, as expected. Similarly, 0.9999 represents



limn(910+9102++910n)=9101110=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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