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