Introduction to Mathematical Analysis I - Second Edition

43 From the Monotone Convergence Theorem, we deduce that there is ` ∈ R such that lim n → ∞ a n = ` . Since the subsequence { a k + 1 } ∞ k = 1 also converges to ` , taking limits on both sides of the equation in ( 2.7 ) , we obtain ` = ` + 5 3 . Therefore, 3 ` = ` + 5 and, hence, ` = 5 / 2. Example 2.3.3 —The number e. Consider the sequence { a n } given by a n = 1 + 1 n n , n ∈ N . By the binomial theorem , a n = n ∑ k = 0 n k 1 n k = 1 + 1 + n ( n − 1 ) 2! 1 n 2 + n ( n − 1 )( n − 2 ) 3! 1 n 3 + · · · + n ( n − 1 ) · · · ( n − ( n − 1 )) n ! 1 n n = 1 + 1 + 1 2! 1 − 1 n + 1 3! 1 − 1 n 1 − 2 n + · · · + 1 n ! 1 − 1 n 1 − 2 n · · · 1 − n − 1 n . The corresponding expression for a n + 1 has one more term and each factor ( 1 − k n ) is replaced by the larger factor ( 1 − k n + 1 ) . It is then clear that a n < a n + 1 for all n ∈ N . Thus, the sequence is increasing. Moreover, a n ≤ 1 + 1 + 1 2! + 1 3! + · · · + 1 n ! < 2 + 1 1 . 2 + 1 2 . 3 + · · · + 1 ( n − 1 ) · n = 2 + n − 1 ∑ k = 1 1 k − 1 k + 1 = 3 − 1 n < 3 . Hence the sequence is bounded above. By the Monotone Convergence Theorem, lim n → ∞ a n exists and is denoted by e . In fact, e is an irrational number and e ≈ 2 . 71828. The following fundamental result is an application of the Monotone Convergence Theorem. Theorem 2.3.3 — Nested Intervals Theorem. Let { I n } ∞ n = 1 be a sequence of nonempty closed bounded intervals satisfying I n + 1 ⊂ I n for all n ∈ N . Then the following hold: (a) T ∞ n = 1 I n 6 = /0. (b) If, in addition, the lengths of the intervals I n converge to zero, then T ∞ n = 1 I n consists of a single point. Proof: Let { I n } be as in the statement with I n = [ a n , b n ] . In particular, a n ≤ b n for all n ∈ N . Given that I n + 1 ⊂ I n , we have a n ≤ a n + 1 and b n + 1 ≤ b n for all n ∈ N . This shows that { a n } is an increasing sequence bounded above by b 1 and { b n } is a decreasing sequence bounded below by a 1 . By the