47 By the binomial theorem, k n n 1 = an ∑ k n k=0 n(n − 1) 1 n(n− 1)(n − 2) 1 n(n − 1)···(n − (n− 1)) 1 = 1 + 1 + ··· + 2! n2 + 3! n3 + n! nn 1 1 1 1 2 1 1 2 n − 1 = 1 + 1 + 1 − + 1 − 1 − + ··· + 1 − 1− ··· 1 − . 2! n 3! n n n! n n n The corresponding expression for an+1 has one more term and each factor (1 − k ) is replaced by the n k larger factor (1− n+1 ). It is then clear that an < an+1 for all n ∈ N. Thus, the sequence is increasing. Moreover, 1 1 1 an ≤ 1 + 1+ + + ··· + 2! 3! n! 1 1 1 < 2+ + + ··· + 1.2 2.3 (n− 1) · n n−1 1 1 1 = 2 + ∑ − = 3 − < 3. k k + 1 n k=1 Hence the sequence is bounded above. By the Monotone Convergence Theorem (Theorem 2.3.1), limn →∞ an 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 2.3.2 — Nested Intervals Theorem. Let {In} ∞ n=1 be a sequence of nonempty closed bounded intervals satisfying In+1 ⊂ In for all n ∈ N. Then the following hold: T∞ (i) n=1 In ̸ = 0/ . T∞ (ii) If, in addition, the lengths of the intervals I n converge to zero, then n=1 In consists of a single point. Proof: Let {In} be as in the statement with In =[an,bn]. In particular, an ≤ bn for all n ∈ N. Given that In+1 ⊂ In, we have an ≤ an+1 and bn+1 ≤ bn for all n ∈ N. This shows that {an} is an increasing sequence bounded above by b1 and {bn} is a decreasing sequence bounded below by a1. By the Monotone Convergence Theorem (Theorem 2.3.1), there exist a,b ∈ R such that limn →∞ an = a and limn →∞ bn = b. Since an ≤ bn for all n, by Theorem 2.1.2, we get a ≤ b. Now, we also have an ≤ a and b ≤ bn for all n ∈ N (since {an} is increasing and {bn} is decreasing). This shows that if T∞ T∞ a ≤ x ≤ b, then x ∈ In for all n ∈ N. Thus, [a,b] ⊂ n=1 In. It follows that n=1 In ̸ = 0/. This proves part (i). T∞ T∞ Now note also that n=1 In ⊂ [a,b]. Indeed, if x ∈ n=1 In, then x ∈ In for all n. Therefore, an ≤ x ≤ bn for all n. Using Theorem 2.1.2, we conclude a ≤ x ≤ b. Thus, x ∈ [a,b]. This proves the T∞ desired inclusion and, hence, n=1 In =[a,b]. We now prove part (ii). Suppose the lengths of the intervals In converge to zero. This means limn →∞(bn − an)= 0. Then b = limn →∞ bn = limn →∞[(bn − an)+ an]= limn →∞(bn − an)+ T∞ limn→∞ an = 0+ a = a. It follows that =1 In = {a} as desired. □ n
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