Proof let s n 2 n p n is false to derive a

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Proof. Let S = { n 2 N | P ( n ) is false. } . To derive a contradiction, we suppose that S is not empty. By the Well Ordering Principle (Corollary 1.13), S has a minimal element. Call the minimal element ( k + 1), so ( k + 1) 2 S N but for any n k , n 62 S . Therefore, P ( k ) is true. But, by (2), P ( k ) true implies P ( k +1) is also true, and this contradicts ( k +1) 2 S . From the proof, we can see that the following version of induction is also verified: If we can show both: (1) P (1) is true; and (2) For every k 2 N , if P ( n ) is assumed to be true for all n k , then P ( k + 1) is true, then P ( n ) is true for all n 2 N . Sometimes this form of induction is needed (for example, for sequences defined by iteration involving several previous terms.) This is discussed in section 1.2 of the textbook. 22
2.3 Divergent Sequences and Subsequences Although we prefer sequences which converge, there are also many interesting things to learn about divergent sequences. Sequences can diverge in various ways, some more interesting than others. The simplest kind of divergence is called proper divergence in the book. Definition 2.15 (Properly divergent sequences) . Let ( x n ) n 2 N be a sequence in R . (a) We say the sequence properly diverges to + 1 if: 8 a > 0 9 K 2 N so that x n > a 8 n K. We write lim n !1 x n = + 1 or x n ---! n !1 + 1 , even though the sequence does not converge. (b) We say the sequence properly diverges to -1 if: 8 b < 0 9 K 2 N so that x n < b 8 n K. We write lim n !1 x n = -1 or x n ---! n !1 -1 , even though the sequence does not converge. So a sequence diverges to infinity if, for any fixed number a > 0, x n is eventually always larger than a . Example 2.16. Let x n = p n (3 + sin( n )), n 2 N . Then x n ---! n !1 + 1 . Choose any a > 0. Since sin( n ) ≥ - 1 for any n , we have x n p n (3 - 1) = 2 p n > a provided n > a 2 4 . So, take K 2 N with K > a 2 4 . If n K > a 2 4 , by the above calculation, x n > a , and by definition x n ---! n !1 + 1 . Proposition 2.17. If ( x n ) n 2 N is monotone increasing and not bounded above, then x n ---! n !1 + 1 (properly divergent). If ( x n ) n 2 N is monotone decreasing and not bounded below, then x n ---! n !1 -1 (properly divergent). Proof. Take any a > 0. If ( x n ) n 2 N is unbounded above, then in particular a is not an upper bound for the sequence, so there exists K 2 N so that a < x K . Since the sequence is monotone increasing, x K x n for all n K , and therefore a < x K < x n 8 n K. By definition, x n ---! n !1 + 1 . The second statement is similar, and is left as an exercise. 23 K is a natural every a o there exists ke natural so that k da every n K Xa a Xn Lrn a n I 4 n K I I
Example 2.18. Let r > 1 and define the sequence x n = r n , n 2 N . Then, x n +1 = r n +1 = r x n > x n 8 n 2 N , so ( x n ) n 2 N is (strictly) monotone increasing. In particular, notice that x n > 1 8 n . We claim that the sequence is not bounded above, in which case we can apply Propo- sition 2.17 to conclude x n ---! n !1 + 1 . We argue by contradiction, and suppose that ( x n ) is bounded above. In that case, by the Monotone Sequence Theorem, x n = r n ---! n !1 x for some x 2 R . But we pass to the limit in the equation x n +1 = r x n , to obtain the equation x = r x . Since r > 1, we must have x = 0. But, by Theorem 2.4, x 1, a contradiction.

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