On-line Math 21

Example 1 {c_n} = {3, 3.1, 3.14, 3.141, 3.1415, 3.14159,¼} (This should remind you of something).

[Solution]This one isn't quite fair, since I don't tell you a specific formula to compute c_n for each n . In exams like the SAT, this sort of description of a sequence is common, but not so in calculus. Without a formula, there is no reason to suspect that the next term is not 37. However, we are pretty good at pattern recognition, and you probably see that c_n is supposed to be p to n significant figures. Actually, the whole concept of infinite decimal expansions is really the same as a sequence (and can also be expressed as a series), where the final number is the limit of the process of taking more and more terms of the decimal expansion. So,

lim n® ¥  cn = p.

Of course, there is a way to write a formula for c_n , it just isn't paticularly helpful. One way to do that would be to say that c_n = [10^n-1p]/10^n-1 , where [x] is called the ``greatest integer'' of x , for example, [2.1] = 2 .

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Copyright (c) 2000 by David L. Johnson.