Introduction to Mathematical Analysis I - 3rd Edition

47 By the binomial theorem, an = n ∑ k=0 n k 1 n k = 1+1+ n(n−1) 2! 1 n2 + n(n−1)(n−2) 3! 1 n3 +· · ·+ n(n−1)· · ·(n−(n−1)) n! 1 nn = 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 an+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 an <an+1 for all n∈N. Thus, the sequence is increasing. Moreover, an ≤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 (Theorem 2.3.1), limn→∞an exists and is denoted by e. In fact, e is an irrational number ande ≈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: (i) T∞ n=1In̸ =0/ . (ii) If, in addition, the lengths of the intervals In converge to zero, then T ∞ n=1In 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 ∈Rsuch 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 ≤aand b≤bn for all n∈N(since {an}is increasing and {bn}is decreasing). This shows that if a≤x ≤b, thenx ∈In for all n∈N. Thus, [a,b] ⊂T ∞ n=1In. It follows that T ∞ n=1In̸ =0/ . This proves part (i). Now note also that T∞ n=1In ⊂[a,b]. Indeed, if x ∈T ∞ n=1In, 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 desired inclusion and, hence, T∞ n=1In = [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) + limn→∞an =0+a=a. It follows that T ∞ n=1In ={a}as desired. □

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