55 ■ Example 2.5.1 Consider the sequence {an} given byan =(−1) n. For any n∈N, sn =sup{ak : k ≥n}=1and tn =inf{ak : k ≥n}=−1. Then limn →∞ sn =1 and limn →∞tn =−1. Thus, lim supn →∞ an =1 and lim infn →∞ an =−1. ■ Example 2.5.2 Consider the sequence {an} given byan =(−1) nn. For any n∈N, sn =sup{ak : k ≥n}=∞and tn =inf{ak : k ≥n}=−∞. Then limn →∞ sn =∞and limn →∞tn =−∞. Thus, lim supn →∞ an =∞and liminfn →∞ an =−∞. ■ Example 2.5.3 Consider the sequence {an} given byan =n. For any n∈N, sn =sup{ak : k ≥n}=∞and tn =inf{ak : k ≥n}=n. Then limn →∞ sn =∞and limn →∞tn =limn →∞ n=∞. Thus, lim supn →∞ an =∞and liminfn →∞ an =∞. In a similar way, if {bn}is givenbybn =−n, we have limsupn →∞ bn =−∞and liminfn →∞ bn =−∞. Proposition 2.5.1 Let {an} be a bounded sequence of real numbers. Then limsupn →∞ an and liminfn →∞ an exist (as real numbers). Proof: Since {an}is bounded, we see that the sequence {sn}defined in (2.8) is a bounded sequence of real numbers, so it is bounded below. If m≤n, then {ak : k ≥n}⊂{ak : k ≥m}. Thus, it follows from Theorem 1.5.3 that sn ≤sm, so the sequence {sn}is decreasing. Similarly, the sequence {tn} defined in (2.9) is increasing and bounded above. Therefore, both sequences are convergent by Theorem 2.3.1. By the definition, lim supn →∞ an and liminfn →∞ an exist (as real numbers). □ Proposition 2.5.2 Let {an} be a sequence of real numbers. If {an} is not bounded above, then limsup n→∞ an =∞. Similarly, if {an} is not bounded below, then liminf n→∞ an =−∞, Proof: Suppose {an} is not bounded above. We will show that limn →∞ sn =∞, where sn is defined in (2.8). Since {an} is not bounded above, for any n∈N, the set {ak : k ≥n} is also not bounded above. Thus, sn =sup{ak : k ≥n}=∞for all n. Therefore, limsup n→∞ an =lim n→∞ sn =∞. The proof for the second case is similar. □ Proposition 2.5.3 Let {an} be a sequence of real numbers. Then (i) limsupn →∞ an =−∞if and only if limn →∞ an =−∞, (ii) lim infn →∞ an =∞if and only if limn →∞ an =∞.
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