46 2.3 Monotone Sequences ■ Example 2.3.1 Given r ∈Rwith|r| <1, define an =r n for n∈N. Then lim n→∞ an =0. This is clear if r =0. Let us first consider the case where 0<r <1. Then 0≤an+1 =ran ≤an for all n. Therefore, {an}is decreasing and bounded below. By Theorem 2.3.1, the sequence converges. Let ℓ =lim n→∞ an. Since an+1 =ran for all n, taking limits on both sides gives ℓ =rℓ. Thus, (1−r)ℓ =0 and, hence, ℓ =0. In the general case, we only need to consider the sequence defined bybn =|an| for n∈N; see Exercise 2.1.4. ■ Example 2.3.2 Consider the sequence {an}defined as follows: a1 =2 (2.6) an+1 = an +5 3 for n≥1. (2.7) We will apply the Monotone Convergence Theorem (Theorem 2.3.1) to show that the sequence converges and then find the value of its limit. First we will show that the sequence is increasing. We prove by induction that for all n∈N, an <an+1. Since a2 = a1+5 3 = 7 3 >2=a1, the statement is true for n=1. Next, suppose ak <ak+1 for some k ∈N. Thenak +5<ak+1 +5 and(ak +5)/3< (ak+1 +5)/3. Therefore, ak+1 = ak +5 3 < ak+1 +5 3 =ak+2. It follows by induction that the sequence is increasing. Next we prove that the sequence is bounded above by 3. Again, we proceed by induction. The statement is clearly true for n=1. Suppose that ak ≤3 for some k ∈N. Then ak+1 = ak +5 3 ≤ 3+5 3 = 8 3 ≤ 3. It follows that an ≤3 for all n∈N. From the Monotone Convergence Theorem (Theorem 2.3.1), we deduce that there is anℓ ∈R such that limn→∞an =ℓ. Since the subsequence {ak+1} ∞ k=1 also converges to ℓ, taking limits on both sides of the equation in (2.7), we obtain ℓ = ℓ+5 3 . Therefore, 3ℓ =ℓ+5 and, hence, ℓ =5/2. ■ Example 2.3.3 —The number e. Consider the sequence {an}given by an = 1+ 1 n n , n∈N.
RkJQdWJsaXNoZXIy NTc4NTAz