If s n does not converge to any real number then we say that it diverges The

If s n does not converge to any real number then we

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If ( s n ) does not converge to any real number then we say that it diverges . The simplest example of a sequence is a constant sequence . If C is a con- stant, let s 1 = C, s 2 = C, s 3 = C, · · · . It is clear that this sequence has limit C too, since for any ǫ > 0, | s n C | = 0 < ǫ n 1. Another simple example from calculus is the sequence ( 1 n ). This converges to 0 since given ǫ > 0, choose N > 1 ǫ , then n N ⇒ | 1 n 0 | = 1 n 1 N < ǫ . A sequence ( s n ) is called bounded if the set { s n : n N } is a bounded set. That is, there are numbers m and M such that m s n M for all n N . This is the same as saying that { s n : n N } ⊂ [ m, M ]. It is easy to see that this is equivalent to: there exists a number K 0 such that | s n | ≤ K for all n N . (See the first lines of the last Section.) Fact 1. Any convergent sequence is bounded. Proof: Suppose that s n s as n → ∞ . Taking ǫ = 1 in the definition of convergence gives that there exists a number N N such that | s n s | < 1 whenever n N . Thus | s n | = | s n s + s | ≤ | s n s | + | s | < 1 + | s | whenever n N . Now let M = max {| s 1 | , | s 2 | , · · · , | s N | , 1+ | s |} . We have | s n | ≤ M if n = 1 , 2 , · · · , N , and | s n | ≤ M if n N . So ( s n ) is bounded. A sequence ( a n ) is called nonnegative if a n 0 for all n N . To say that a nonnegative sequence converges to zero is simply to say that: ǫ > 0 , Ns.t. a n < ǫ n N. This is because | a n s | = | a n 0 | = | a n | = a n in this case. The reason we introduce nonnegative sequences, is that the convergence of a general sequence can be restated as another nonnegative sequence converging to 0. In fact:
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34 MATH 3333–INTERMEDIATE ANALYSIS–BLECHER NOTES Fact 2. If ( s n ) is a general sequence then: lim n s n = s ⇐⇒ lim n ( s n s ) = 0 ⇐⇒ lim n | s n s | = 0 . That is, the sequence ( s n ) converges to s if and only if the nonnegative sequence ( | s n s | ) converges to 0. To prove this, simply look at the ǫ -definition of lim n s n = s , and compare it with the ǫ -definition of lim n | s n s | = 0. Next we will prove some facts for nonnegative sequences, which we will use later, together with Fact 2, to prove many other facts about general sequences. Fact 3. If ( a n ) and ( b n ) are nonnegative sequences, with lim n a n = 0 and lim n b n = 0, and if C 0, then lim n a n + b n = lim n Ca n = 0 . Also, if lim n a n = 0 and if ( b n ) is any bounded sequence, then lim n a n b n = 0. Proof: Let ǫ > 0 be given. Then N 1 such that a n < ǫ whenever n N 1 . In fact, replacing ǫ by ǫ/ 2, we can say that N 1 such that a n < ǫ/ 2 whenever n N 1 . Similarly, N 2 such that b n < ǫ/ 2 whenever n N 2 . Let N = max { N 1 , N 2 } , so that if n N then n N 1 and n N 2 . Thus if n N then a n + b n < ǫ/ 2+ ǫ/ 2 = ǫ . We have proved that lim n a n + b n = 0. To prove lim n Ca n = 0, we can assume that C > 0, because if C = 0 this result is obvious. If ǫ > 0 is given, then replacing ǫ by ǫ/C , we can say that N 3 such that a n < ǫ/C whenever n N 3 . Thus Ca n < ǫ whenever n N 3 . Hence lim n Ca n = 0 .
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