34 2.1 Convergence Remark 2.1.1 It follows directly from the definition, using the Archimedean property, that a sequence {an}converges to aif and only if for any ε >0, there exists a real number Nsuch that |an −a| <ε for any n∈Nwithn>N. Remark 2.1.2 Applying Proposition 1.4.2(iv), the condition |an −a| <ε in Definition 2.1.1 is equivalent to a−ε <an <a+ε for any n∈Nwithn≥N or an ∈(a−ε,a+ε) for any n∈Nwithn≥N. ■ Example 2.1.1 Let {an}be the sequence given byan = 1 n for n∈N. We claim that limn→∞an =0. First notice that in this example a=0. We verify the claim using the definition. Given ε >0, we seek an N∈Nsuch that if n≥N then |an −a| <ε. To find a suitable positive integer N we start with the expression|an −a|. We have |an −a| = 1 n− 0 = 1 n = 1 n . We look for N∈Nsuch that if n≥N, then 1 n <ε. This suggests that N>1/ε is a good choice since then 1 n ≤ 1 N < 1 1/ε =ε. Now that we have found N, the last step is to write a formal proof. Let ε >0 and choose an integer N, N>1/ε. Note that such a positive integer Nexists due to the Archimedean property (Theorem 1.6.1). If n∈Nwithn≥N, we have |an −a| = 1 n− 0 = 1 n ≤ 1 N < 1 1/ε =ε. This shows that limn→∞1/n=0. ■ Example 2.1.2 We now generalize the previous example as follows. Let α>0 and consider the sequence {an}given by an = 1 nα for n∈N. We will show that limn→∞an =0. Let ε >0. As in the previous example, we start with the expression|an −a| and find a suitable condition for N∈N. We have |an −a| = 1 nα − 0 = 1 nα . Observe that if n≥N, then 1 nα ≤ 1 Nα . We are seekingNsuch that 1 Nα <ε. Solving the inequality for Ngives N>( 1 ε ) 1/α. The last step is to write a formal proof. Let ε >0 and choose an integer Nsatisfying N>(1 ε ) 1/α. Such an Nexists by the Archimedean property (Theorem 1.6.1). For everyn≥N, one has nα ≥Nα. This implies 1 nα − 0 = 1 nα ≤ 1 Nα < 1 1/ε =ε. We conclude that limn→∞1/n α =0.
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