,... We would like to compare this with the sequence y n = 1 n

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1 Example 2.0 Let (x n ) n= be the sequence given by x n = 2, i.e. n 2, 4, 8, 6,.... We would like to compare this with the sequence = n (which we know converges to zero). We claim that 2 n n, n N. Proof. We prove this by induction. We know that the statement is true for n = (where it says that 2 ). Now assume that the claim holds for some n N. Then 2 n+ = 2 2 n ( ) 2 n = n + n n +. So the fact that the claim holds for n [we used this in ( )] implies that it also holds for n +. Hence it holds for all n N. The claim, together with Lemma 2.3, then implies that (x n ) converges to zero. We want to generalise this and show that the sequence given by x n = c n for some c R converges to zero if c < (and does not converge to zero if c ). To prove this, it is useful to prove the following lemma first. Lemma 2. (Bernoulli s inequality) If α R, α, then n N : ( + α) n + nα. Proof. We prove this again by induction. Obviously the statement is true for n = (where it just says + α + α). Assume now that it is true for some n N. Then ( + α) n+ = ( + α)( + α) n ( ) ( + α)( + nα) = + α + nα + nα 2 + (n + )α, where in ( ) we used the induction hypothesis as well as α. Hence the claim being true for n implies that it is true for n + as well. Thus it holds for all n N. Theorem 2.2 ( Geometric sequences ) If c R and the sequence (x n ) n= is defined by x n = c n, then i) (x n ) converges to zero if c <. ii) (x n ) does not converge to zero if c. Proof. i) If c <, then c = +α for some α > 0. So by Lemma 2., we have c n = c n = ( ) n + α + nα < nα. But ( nα ) converges to zero (by Lemma 2.5, since n converges to zero). Hence the sequence (c n ) converges to zero by Lemma

2 ii) If c then c = + α for some α 0. Hence, by Lemma 2. c n = c n + nα, for all n N. But the constant sequence,,,... does not converge to zero and hence (c n ) does not converge to zero by Corollary 2.4. What about products and quotients of sequences? If (x n ) n= and () n= both converge to zero, what can we say about the sequences (x n ) xn n= and ( ) n=? We cannot say much about the quotient ( xn ) n=. It will depend on whether the nominator or denominator go to zero faster. For example, let (x n ) n= be given by x n = n and () n= defined by =. We know that (x n 2 n ) and ( ) both converge to zero. But we have: xn yn x n = n, so ( xn ) does not converge to zero. = yn n, so ( x n ) does converge to zero. Differently from this fact, we can always say something about the product (x n ) n=. Theorem 2.3 (Product of sequences converging to zero) If (x n ) n= and () n= both converge to zero then (x n ) n= converges to zero. Proof. We follow a similar strategy to the proof of Theorem 2.8, but instead of splitting ε into ε 2 + ε 2, we split it into ε ε. We must prove that ε > 0 N N n N, n > N : x n < ε. To prove this, suppose we are given ε > 0. Then N x N n N, n > N x : x n < ε = ε since (x n ) converges to zero, and N y N n N, n > N y : < ε = ε since ( ) converges to zero. Let N = max{n x, N y }. Then for all n > N, we have x n = x n < ε ε = ε. Hence (x n ) n= converges to zero. Example 2.4 Consider x n = 2 n 3 n (n+) 2. By Lemma 2.3, ( (n+) 2 ) converges to zero, as (n+) 2 < n. By Theorem 2.2, 2 n 3 n = ( 2 3 )n converges to zero. Thus, by Theorem 2.3, (x n ) converges to zero. 27

3 Remark. Of course, we could have obtained this also by estimating x n < n and using only earlier results, but sometimes it is difficult to obtain such estimates. The proof of Theorem 2.3 suggests that we might be able to prove more general results by splitting ε up in other ways, e.g. ε = ε c c for some real constant c > 0. This is our next goal. Definition 2.5 (bounded sequences) We say that a sequence (x n ) n= (of real numbers) is bounded above if M R such that x n M for all n N. The sequence is bounded below if m R such that x n m for all n N. We say that (x n ) is bounded if it is bounded above and below. M R m N Remark. Clearly, (x n ) n= is bounded if and only if ( x n ) n= is bounded above. Example 2.6 Every sequence (x n ) n= which converges to zero is bounded. Proof. If (x n ) converges to zero then taking ε = in the definition of convergence to zero (Definition 2.), there exists some N N such that for all n > N we have x n <. This is illustrated in the following figure. R N 28

4 Take M to be max{ x, x 2,..., x N, }. Then for all n N, we have x n M, i.e. M x n M for all n N. This shows that M is an upper bound and m = M is a lower bound for the sequence (x n ) as defined in Definition 2.5. We are now ready to state and prove the more general version of Theorem 2.3. Theorem 2.7 (Product of sequences, more general version) If (x n ) n= is bounded and () n= converges to zero then (x n ) n= zero. converges to Proof. As (x n ) n= is bounded, there exists M R such that x n M for all n N. Hence x n = x n M = M, n N. But ( ) converges to zero and so (M ) converges to zero as well (by Lemma 2.5), so (x n ) converges to zero by Lemma 2.3. Remark. Alternatively, we could prove this like we proved Theorem 2.3, splitting up ε = M ε M. (As () converges to zero, we will get ε M for large enough n.) 2.B Convergence of sequences to non-zero elements of R Definition 2.8 (Convergence of a sequence to x R) A sequence (x n ) n= converges to x R if and only if (x n x) n= zero. Equivalently, (x n ) n= converges to x R if and only if converges to ε > 0 N N n > N : x n x < ε. (2.4) We use the notation x n x as n or lim n x n = x. Theorem 2.9 Suppose (x n ) n= converges to x R and () n= converges to y R. Then i) (cx n ) n= converges to cx for any constant c R. ii) (x n + ) n= converges to x + y. iii) (x n ) n= converges to xy. iv) if y 0 and 0 for all n N, then ( xn ) n= converges to x y. Proof. i) By the definition of convergence of (x n ) to x, we know that the sequence (x n x) converges to zero. So by Lemma 2.5, the sequence c(x n x) converges to zero, i.e. (cx n cx) converges to zero. But then by definition (cx n ) converges to cx. 29

5 ii) As x n x and y as n, we know that (x n x) and ( y) are two sequences that converge to zero. Hence by Theorem 2.8, ((x n x)+( y)) n= converges to zero. Rewriting this, we obtain that ((x n + ) (x + y)) n= converges to zero. But then by definition (x n + ) converges to x + y. iii) We know that (x n x) and ( y) converge to zero. We write Then, we know that x n xy = (x n x)( y) + x( y) + (x n x)y. (2.5) ((x n y)( y)) converges to zero (by Theorem 2.3). (x( y)) converges to zero (by Lemma 2.5, as ( y) converges to zero). ((x n x)y) converges to zero (again by Lemma 2.5, as (x n x) converges to zero). So by Theorem 2.8, the right hand side of (2.5) converges to zero and hence (x n xy) converges to zero. But then by definition (x n ) converges to xy. iv) This is a question on Coursework Sheet 4. Examples 2.20 i) Consider the sequence (x n ) given by x n = c for all n N. This sequence converges to c. Proof. We must prove that (x n c) converges to zero. But x n c = 0 for all n N and we already know that the zero sequence converges to zero. ii) x n = 3n2 +5n+2 2n 2 +n+. This sequence converges to 3 2. Proof. We write x n in the form x n = n + 2 n n + n 2. But (3 + 5 n + 2 n 2 ) converges to = 3. [This uses Theorem 2.9 ii) and the facts that the constant sequence 3 converges to 3 while ( 5 n ) and ( 2 n 2 ) converge to zero the latter due to Lemma 2.3 and Lemma 2.5, as seen several times before.] Similarly, (2 + n + n 2 ) converges to = 2, again by Theorem 2.9 ii) [and the fact that ( n ) and ( n 2 ) converge to zero]. In a last step, we then apply Theorem 2.9 iv) to conclude that (x n ) converges to

6 Lemma 2.2 (Limits are unique) If (x n ) n= converges to x R and also converges to y R then x = y. That is: if a limit exists, then it is unique. Proof. Suppose x y. Without loss of generality, assume y > x and write α for the difference y x = y x. Since (x n ) converges to x, we know that N x N such that n > N x, we have x n x < α 2. Similarly, since (x n ) converges to y, we know that N y N such that n > N y, we have x n y < α 2. Now consider some x n with n max{n x, N y }. This x n satisfies x n x < α 2 and x n y < α 2. Hence, using the triangle inequality, y x y x n + x n x = x n y + x n x < α 2 + α 2 = α. So y x < α, contradicting our definition of α as x y. Lemma 2.22 Suppose we have two sequences (x n ) and ( ) with x n x and y as n. Suppose further that x n for all n N. Then x y. Proof. Since x n x as n, we can find for every ε > 0 some N x N such that n > N x : x n x < ε, hence in particular x n > x ε. Since y as n, we can find for every ε > 0 some N y N such that n > N y : y < ε, and thus in particular < y + ε. Setting N = max{n x, N y } we obtain for all n > N that x ε < x n < y + ε, that is x y < 2ε. But because this is true for every ε > 0, we conclude that x y 0, or equivalently x y. Examples 2.23 i) If (x n ) n= converges to x R and there exists a constant c R such that x n c for all n N, then x c. Proof. We know that ( ) defined by = c for all n N converges to c. We can then apply Lemma 2.22 to (x n ) and this choice of ( ). Similarly, if (x n ) n= converges to x R and there exists a constant c R such that x n c for all n N, then x c. 3

7 ii) If the inequality x n in Lemma 2.22 is replaced by a strict inequality x n <, we still can only conclude x y and not x < y. An example for this is x n = 0 and = n for all n N. Although we have x n < for all n N, both sequences converge to zero, so x = lim n x n = 0 and y = lim n = 0 and we have x = y. iii) The sandwich principle : If ( ) is some sequence and (x n ) and (z n ) are two converging sequences such that x n z n and lim n x n = lim n z n, then ( ) converges as well and lim n = lim n x n. Proof. This is a question on Coursework Sheet 4. 2.C Monotonic sequences Definition 2.24 (Monotonic sequences) Let (x n ) n= be a sequence. We say (x n ) is an increasing sequence if x n+ x n for all n N. We say (x n ) is strictly increasing if x n+ > x n for all n N. We say (x n ) is a decreasing sequence if x n+ x n for all n N. We say (x n ) is strictly decreasing if x n+ < x n for all n N. We say (x n ) is monotonic if it is either increasing or decreasing. Examples 2.25 i) x n = n is strictly decreasing since n+ < n for all n N. Similarly, x n = 3+ n is strictly decreasing as well. ii) x n = n or x n = 2 n are strictly increasing. iii) x n = for all n N is both increasing and decreasing (but not strictly). iv) x n = n is strictly increasing. v) x n = ( )n n is neither increasing nor decreasing. The sequences in the examples i) iii) are monotonic and bounded. And all of them converge to a limit in R. Example iv) is monotonic but not bounded and it does not converge to a limit in R. The next theorem says that this is not just a coincidence. 32