Convergent Sequences PDF

Convergent Sequences Definition 1. A sequence of real numbers (sn ) is said to converge to a real number s if ∀ε > 0, ∃N ∈ N, such that n > N implies |sn − s| < ε. (1) When this holds, we say that (sn ) is a convergence sequence with s being its limit, and...

Convergent Sequences Definition 1. A sequence of real numbers (sn ) is said to converge to a real number s if ∀ε > 0, ∃N ∈ N, such that n > N implies |sn − s| < ε. (1) When this holds, we say that (sn ) is a convergence sequence with s being its limit, and write sn → s or s = limn→∞ sn. If (sn ) does not converge, then we say that (sn ) is a divergent sequence. We first show that one sequence (sn ) can not have two different limits. Suppose sn → s and sn → t. Let ε > 0. Then 2ε > 0. Since sn → s, by definition there is N1 ∈ N such that for n > N1 , |sn − s| < 2ε. Since sn → t, by definition there is N2 ∈ N such that for n > N2 , |sn − t| < 2ε. Here we use N1 and N2 in the two statements because the N coming from the two limits may not be the same. Let N = max{N1 , N2 }. If n > N , then n > N1 and n > N2 both hold. So |sn − s| < 2ε and |sn − t| < 2ε , which by triangle inequality imply that ε ε |s − t| ≤ |sn − s| + |sn − t| < + = ε. 2 2 Now |s−t| < ε holds for every ε > 0. We then conclude that |s−t| = 0 (for otherwise |s−t| > 0, we then get a contradiction by choosing ε = |s − t|). So s = t, and the uniqueness holds. We will use the following tools to check whether a sequence converges or diverges. 1. the definition 2. basic examples 3. limit theorems 4. boundedness and subsequences. We have stated the definition. Now we consider some examples. Example 1. Let s ∈ R. If sn = s for all n, i.e., (sn ) is a constant sequence, then lim sn = s. Proof. For any given ε > 0 we simply choose N = 1. If n > N , then |sn − s| = 0 < ε. 1 Example 2. We have n → 0. 1 Proof. Let ε > 0. By Archimedean property, there is N ∈ N such that N < ε. If n > N , then 1 1 1 | − 0| = < < ε. n n N Example 3. The following two sequences are divergent 1  (i) (sn ) = ((−1)n ) = (−1, 1, −1, 1, −1, 1,... ); (ii) (sn ) = (n) = (1, 2, 3, 4, 5, 6,... ). Proof. (i) We use the notation of subsequence and statement that will be proved later. Suppose n1 < n2 < n3 < · · · is a strictly increasing sequence of indices, then (snk ) is a subsequence of (sn ). We will prove a theorem, which asserts that, if (sn ) converges to s, then any subsequence of (sn ) also converges to s. The sequence (sn ) = ((−1)n ) contains two constant sequences (1, 1, 1,... ) (with nk = 2k) and (−1, −1, −1,... ) (with nk = 2k −1), which converge to different limits. So the original (sn ) can not converge. (ii) We use the following theorem. If (sn ) is convergent, then it is a bounded sequence. In other words, the set {sn : n ∈ N} is bounded. So an unbounded sequence must diverge. Since for sn = n, n ∈ N, the set {sn : n ∈ N} = N is unbounded, the sequence (n) is divergent. Remark 1. This example shows that we have two ways to prove that a sequence is divergent: (i) find two subsequences that convergent to different limits; (ii) show that the sequence is unbounded. Note that the (sn ) in (i) is bounded and divergent. The (sn ) in (ii) is divergent, but lim sn actually exists, which is +∞, and its every subsequence also tends to +∞. We will define that limit later. Now we state some limit theorems. Theorem 1 (Theorem 9.1). Every convergent sequence is bounded. Proof. Let (sn ) be a sequence that converges to s ∈ R. Applying the definition to ε = 1, we see that there is N ∈ N such that for any n > N , |sn − s| < 1, which then implies that |sn | ≤ |s| + 1. Let M = max{|s1 |, |s2 |,... , |sN |, |s| + 1}. The maximum exists since the set is finite. Then for any n ∈ N, |sn | ≤ M (consider the case n ≤ N and n > N separately), i.e., −M ≤ sn ≤ M. So {sn : n ∈ N} is bounded. Theorem 2 (Theorem 9.3). If (sn ) converges to s and (tn ) converges to t, then (sn + tn ) converges to s + t. Proof. Let ε > 0. Then 2ε > 0. Since sn → s, there is N1 ∈ N such that for n > N1 , |sn −s| < 2ε. Since tn → t, there is N2 ∈ N such that for n > N2 , |tn − t| < 2ε. Let N = max{N1 , N2 }. If n > N , then n > N1 and n > N2 both hold, and so |sn − s| < 2ε and |tn − t| < 2ε , which together imply (by triangle inequality) that ε ε |(sn + tn ) − (s + t)| ≤ |sn − s| + |tn − t| < + = ε. 2 2 So we have the desired convergence. Theorem 3 (Theorem 9.4). If (sn ) converges to s and (tn ) converges to t, then (sn ·tn ) converges to s · t. 2 Discussion. We need to bound |sn tn − st| from above for big n. We write sn tn − st = sn tn − sn t + sn t − st = sn (tn − t) + t(sn − s). By triangle inequality, we get |sn tn − st| ≤ |sn (tn − t)| + |t(sn − s)| = |sn ||tn − t| + |t||sn − s|. Since tn → t and sn → s, we know that |tn − t| and |sn − s| can be arbitrarily small if we choose n big enough. Thus, if |sn | and |t| are not too big, then we can control the sum on the RHS (righthand side). In fact, the size of |sn | can be controlled because of Theorem 9.1. Proof. Since (sn ) is convergent, by Theorem 9.1, there is M > 0 such that |sn | ≤ M for every ε n. We may choose M big such that M ≥ |t|. Let ε > 0. Then 2M > 0. Since sn → s, there ε is N1 ∈ N such that for n > N1 , |sn − s| < 2M. Since tn → t, there is N2 ∈ N such that for ε n > N2 , |tn − t| < 2M. Let N = max{N1 , N2 }. If n > N , then n > N1 and n > N2 both hold, ε ε and so |sn − s| < 2M and |tn − t| < 2M , which together with |sn | ≤ M and |t| ≤ M imply that |sn tn − st| ≤ |sn (tn − t)| + |t(sn − s)| = |sn ||tn − t| + |t||sn − s| ε ε ≤ M |tn − t| + M |sn − s| < M +M = ε. 2M 2M Corollary 1. If (sn ) converges to s, k ∈ R, and m ∈ N, then (ksn ) converges to ks and sm n converges to sm. Proof. For the sequence (ksn ), we apply Theorem 9.4 to the sequence (tn ) with tn = k for all n. For the sequence (sm n ) we use induction. In the induction step, note that sn m+1 = s ∗ sm n n and apply Theorem 9.4 to tn = sn m Corollary 2. If (sn ) converges to s and (tn ) converges to t, then (sn − tn ) converges to s − t. Proof. We write sn + tn = sn + (−1)tn and apply Theorem 9.3 and the previous corollary. From this corollary we see that sn → s iff sn − s → 0. By the Theorem below, the latter statement is equivalent to that |sn − s| → 0. Theorem 4. (a) Suppose two sequences (sn ) and (tn ) satisfy that tn → 0 and |sn | ≤ |tn | for all but finitely many n. Then sn → 0. (b) For any sequence (sn ), sn → 0 if and only if |sn | → 0. Proof. (a) Let N0 ∈ N be such that |sn | ≤ |tn | for n > N0. Let ε > 0. Since tn → 0, there is N1 ∈ N such that for n > N1 , |tn − 0| < ε. Let N = max{N0 , N1 }. For n > N , |sn | ≤ |tn | and |tn − 0| < ε, which imply that |sn − 0| = |sn | ≤ |tn | = |tn − 0| < ε. (b) From (a) we know that if |sn | = |tn | for all n, then sn → 0 iff tn → 0. We then apply this result to tn = |sn | and use that ||sn || = |sn |. 3 Lemma 1 (Lemma 9.5). If (sn ) converges to s such that s 6= 0 and sn 6= 0 for all n, then (1/sn ) converges to 1/s. Discussion. We need to bound |1/sn − 1/s| from above for big n. We write 1 1 s − sn |sn − s| − = =. sn s sn s |sn ||s| Since sn → s, |sn − s| can be arbitrarily small if we choose n big enough. Thus, if |sn | and |s| are not too close to 0, then we can control the size of the RHS. This means that we need a positive lower bound of the set {|s1 |, |s2 |,... }. Proof. Since s 6= 0, we have |s| |s| 2 > 0. Since sn → s, applying the definition to ε = 2 , we get N ∈ N such that for n > N , |sn − s| < |s| 2 , which then implies by triangle inequality that |sn | ≥ |s| − |sn − s| > |s| − 2 = 2. Let m = min{|s1 |, |s2 |,... , |sN |, |s| |s| |s| 2 }. Then m exists and is positive since the set is a finite set of positive numbers. Let ε > 0. Then m|s|ε > 0. Since sn → s, there is N 0 ∈ N such that n > N 0 implies that |sn − s| < m|s|ε, which together with |sn | ≥ m for all n implies that 1 1 |sn − s| |sn − s| m|s|ε − = ≤ < = ε. sn s |sn ||s| m|s| m|s| Theorem 5 (Theorem 9.6). Suppose (sn ) converges to s and (tn ) converges to t. If s 6= 0 and sn 6= 0 for all n, then (tn /sn ) converges to t/s. Proof. By Lemma 9.5, (1/sn ) converges to 1/s. Applying Theorem 9.4 to the sequences (1/sn ) and (tn ), we get the conclusion. 3 Example 4. Derive lim 3n+1 4n +3n 7n−4 and lim n3 −6 Solution. We write 3n + 1 3 + 1/n 4n3 + 3n 4 + 3 ∗ (1/n)2 = , =. 7n − 4 7 + (−4) ∗ 1/n n3 − 6 1 + (−6) ∗ 1/n We have shown that lim 1/n = 0. So (i) lim(3 + 1/n) = 3 + 0 = 3 and lim(7 + (−4) ∗ 1/n) = 7 + (−4) ∗ 0 = 7, which imply that lim 3n+1 7n−4 = lim(3 + 1/n)/ lim(7 + (−4) ∗ 1/n) = 3/7; (ii) lim(4 + 3 ∗ (1/n)2 ) = 4 + 3 ∗ 02 = 4 and lim(1 + (−6) ∗ 1/n) = 1 + (−6) ∗ 0 = 1, which imply 3 that lim 4nn3+3n −6 = lim(4 + 3 ∗ (1/n)2 )/ lim(1 + (−6) ∗ 1/n) = 4. We now state some theorems about the relation between limits and orders. Theorem 6 (Exercise 8.9). (a) If (sn ) converges to s, and there is N0 ∈ N such that sn ≥ 0 for all n > N0 , then s ≥ 0. (b) Suppose (sn ) converges to s and (tn ) converges to t. If there N0 ∈ N such that sn ≤ tn for all n > N0 , then s ≤ t. 4 Proof. (a) We prove by contradiction. Suppose s < 0. Let ε = |s| = −s > 0. Since sn → s, there is N ∈ N such that for n > N , |sn − s| < ε, which implies that sn < s + ε = 0. Let n = max{N, N0 } + 1. Then n > N0 and n > N. From n > N0 we get sn ≥ 0; from n > N we get sn < 0. This is the contradiction. (b) Applying (i) to the sequence (tn − sn ) we conclude that its limit t − s is nonnegative. For x ∈ [0, ∞) and n ∈ N, the power root x1/n is defined as the unique y ∈ [0, ∞) such that yn = x. The uniqueness of such y follows from the fact that if 0 ≤ y1 < y2 , then y1n < y2n. The existence follows from the “Intermediate Value Theorem” for continuous function f (x) = xn , which will be stated and proved later. We now just accept the existence of x1/n for any 1/n 1/n x ∈ [0, ∞). It is clear that 0 ≤ x1 < x2 implies that 0 ≤ x1 < x2. We restrict our attention to [0, ∞) although in the case that n is an odd number, we can also define x1/n for x < 0. √ When n = 2, x1/2 is often written as x. We have the following theorem. √ Theorem 7 (Example 5). Suppose (sn ) converges to s and sn ≥ 0 for all n. Then ( sn ) √ converges to s. √ √ Discussion We want to bound | sn − s| from above for big n. It is useful to note the equality √ √ √ √ √ √ ( sn − s)( sn + s) = ( sn )2 − ( s)2 = sn − s. Taking absolute value, we get √ √ √ √ | sn − s| · | sn + s| = |sn − s|. √ If s > 0, then √ √ |sn − s| |s − s| | sn − s| = √ √ ≤ n√. sn + s s √ √ Proof. By Theorem 6, s ≥ 0. First suppose s > 0. Then s > 0. Let ε > 0. Then sε > 0. √ Since sn → s, there is N ∈ N such that for n > N , |sn − s| < sε, which implies that √ √ √ |sn − s| |sn − s| sε | sn − s| = √ √ ≤ √ < √ = ε. sn + s s s √ We leave the proof in the case s = 0 as an exercise. Note that for x ≥ 0, x < ε iff x2 < ε. 5