Introduction to Mathematical Analysis I - Second Edition

32 2.1 CONVERGENCE Example 2.1.1 Let a n = 1 n for n ∈ N . We claim that lim n → ∞ a n = 0. We verify it using the definition. Let ε > 0. Choose an integer N > 1 / ε . (Note that such an integer N exists due to the Archimidean Property . ) Then, if n ≥ N , we get | a n − 0 | = 1 n = 1 n ≤ 1 N < 1 1 / ε = ε . Example 2.1.2 We now generalize the previous example as follows. Let α > 0 and consider the sequence given by a n = 1 n α for n ∈ N . Let ε > 0. Choose an integer N such that N > ( 1 ε ) 1 / α . For every n ≥ N , one has n > ( 1 ε ) 1 / α and, hence, n α > 1 ε . This implies 1 n α − 0 = 1 n α < 1 1 / ε = ε . We conclude that lim n → ∞ a n = 0. Example 2.1.3 Consider the sequence { a n } where a n = 3 n 2 + 4 2 n 2 + n + 5 . We will prove directly from the definition that this sequence converges to a = 3 2 . Let ε > 0. We first search for a suitable N . To that end, we simplify and estimate the expression | a n − a | . Notice that a n − 3 2 = 3 n 2 + 4 2 n 2 + n + 5 − 3 2 = 2 ( 3 n 2 + 4 ) − 3 ( 2 n 2 + n + 5 ) 2 ( 2 n 2 + n + 5 ) = − 7 − 3 n 2 ( 2 n 2 + n + 5 ) = 3 n + 7 2 ( 2 n 2 + n + 5 ) < 10 n 4 n 2 = 10 4 n . To guarantee that the last expression is less than ε , it will suffice to choose N > 10 4 ε . Indeed, if n ≥ N , we get | a n − a | ≤ 10 4 n ≤ 10 4 N < 10 4 10 4 ε = ε . Example 2.1.4 Let { a n } be given by a n = 4 n 2 − 1 3 n 2 − n . We claim lim n → ∞ ] a n = 4 3 . Let ε > 0. We search for a suitable N . First notice that 4 n 2 − 1 3 n 2 − n − 4 3 = 12 n 2 − 3 − 12 n 2 + 4 n 3 ( 3 n 2 − n ) = 4 n − 3 3 ( 3 n 2 − n ) .