Scalar multiplication and addition of sequences 9

Transcription

1 8 Sequences Proposition. Every subsequence of a convergent sequence (a n ) n N converges to lim n a n. Proof. If (a nk ) k N is a subsequence of (a n ) n N, then n k k for every k. Hence if ε > 0 and N N with a n r < ε for all n N, then also a nk r < ε for all k N. Notice that does not only say that every subsequence of a convergent sequence converges. The statement is that all subsequences of a convergent sequence (a n ) n N converge, and they all converge to the same number, namely to lim n a n. The assertion in is many times used in its contrapositive way. I.e., if we want tocheckthatasequenceisdivergent, wearesometimesabletospotanonconvergent subsequence, or, two subsequences that converge, but not to same the same limit. A prime example is the sequence (a n ) n N = (( 1) n ) n N. You have done question 3 where you have proved from first principles that (( 1) n ) n N is divergent. Using 1.2.7, we can re-confirm this: The subsequence (a 2n ) n N of (a n ) n N is the constant sequence 1, which converges to 1, whereas the subsequence (a 2n+1 ) n N of (a n ) n N is the constant sequence 1, which converges to -1. Hence by 1.2.7, the original sequence (a n ) n N must be divergent.

2 Scalar multiplication and addition of sequences Scalar multiplication and addition of sequences. It would be a tedious undertaking to prove convergence from first principles for all sequences. Instead we are looking for general principles which allow to deduce convergence (and to compute the limits) from known sequences after applying certain operations. In this section we introduce the simplest rules of this sort Scalar multiplication rule. If (a n ) n N is a convergent sequences and c R, then also (c a n ) n N is a convergent sequence and lim (c a n) = c lim a n. n n Proof. A special case here is c = 0. In this case the sequence (c a n ) n N is equal to the constant sequence of value 0, which converges to 0. We may therefore assume that c 0. We write r = lim n a n and we must show that (c a n ) n N converges to c r. According to the definition of convergence we have to pick some real number ε > 0 and we have to find some N N (depending on ε) such that for all n N, ( ) c a n c r < ε. How do we find such an N? Let us first write out what we know by our assumptions on the convergence of the given sequence: As (a n ) n N converges to r, there is some N 1 N (depending on ε) such that for all n N 1, It follows a n r < ε. c a n c r = c (a n r) = c a n r as c 0 < c ε. What now? Recall that we started with ε and we still have to find some N N such that for all n N, c a n c r < ε. On the other hand we were able to find the natural number N 1 (depending on ε) with the property that for all n N 1, c a n c r < c ε. The idea now is to apply the N 1 -argument for ε c instead of ε!! Then the argument above shows that we can find a natural number N 1 with the property that for all n N 1, ε c a n c r < c c = ε. Thus the natural number N, defined as N := N 1 (the symbol := stands for defined as ) that we found for ε c is suitable to solve our initial problem ( ) for ε Sum rule. If (a n ) n N, (b n ) n N are convergent sequences, then also their sum (a n +b n ) n N is a convergent sequence and lim n (a n +b n ) = lim n a n + lim n b n.

3 10 Sequences Proof. We write r = lim n a n, s = lim n b n and we must show that (a n + b n ) n N converges to r+s. According to the definition of convergence we have to pick some real number ε > 0 and we have to find some N N (depending on ε) such that for all n N, ( ) a n +b n (r +s) < ε. How do we find such an N? Let us first write out what we know by our assumptions on the convergence of the two given sequences: As (a n ) n N converges to r, there is some N 1 N (depending on ε) such that for all n N 1, a n r < ε. As (b n ) n N converges to s, there is some N 2 N (depending on ε) such that for all n N 2, b n s < ε. Hence if we take N 3 as the maximum of N 1 and N 2, then we know for all n N 3 : a n r < ε and b n s < ε. Now by the triangle inequality (cf (iii)), we get for all n N 3 : a n +b n (r +s) = (a n r)+(b n s) a n r + b n s ε+ε = 2ε. What now? Recall that we started with ε and we still have to find some N N such that for all n N, a n +b n (r +s) < ε. On the other hand we were able to find the natural number N 3 (depending on ε) with the property that for all n N 3, a n +b n (r +s) < 2ε. The idea now is to apply the N 3 -argument for ε 2 instead of ε!! Hence the argument above shows that we can find a natural number N 3 with the property that for all n N 3, a n +b n (r +s) < 2 ε 2 = ε. Thus the natural number N := N 3 that we found for ε 2 is suitable to solve our initial problem ( ) for ε.

4 Bounded and monotone sequences Bounded and monotone sequences Definition. For a subset S of R and an element r R we write S r if s r for every s S and S < r if s < r for every s S. Similarly the notation r S and r < S has to be understood. (i) Given a subset S of R we call every element r R with S r an upper bound for S. S is called bounded from above if there is an upper bound for S. Similarly every element r R with the property r S is called a lower bound for S. S is called bounded from below if there is a lower bound for S. S is called bounded if S is bounded from above and from below. For example the set of all rational numbers q with q 2 < 2 is bounded from above by 3 2 and bounded from below by 2. (ii) A sequence (a n ) n N is called bounded from above, bounded from below or bounded, if its value set has this property Proposition. Every convergent sequence is bounded. Proof. Let (a n ) n N be our convergent sequence with limit r. We choose ε = 1 and exploit the definition of convergence: Hence we know that there is some N N such that for all n N, a n r < 1; in other words r 1 < a n < r+1. Hence the set of all a n with n N is bounded above by r +1 and bounded below by r 1. However, the set {a 1,...,a N 1 } is finite and therefore bounded. Since the union of two bounded sets is bounded, the value set of (a n ) n N is bounded. The following cannot be proved and is an axiom for real numbers: Completeness axiom of R. Every nonempty subset S of R which has an upper bound, has a least upper bound Rules on least upper bounds and greatest lower bounds. IfS Risboundedfromabove,thenitsleastupperboundiscalledthesupremum of S and denoted by sup(s). For an element x R we have: x = sup(s) S x and for all r < x } there is some {{ s S with r < s }. S r A subset S R is bounded from below if and only if S := { s s S} is bounded from above, because S r r S for all r R. If this is the case, then S has a greatest lower bound, called the infimum of S, denoted by inf(s) and we have inf(s) = sup( S). Observe that the supremum of a bounded set S may be a member of S (e.g. if S = [0,1]) or may not be a member of S (e.g. if S = [0,1)). The existence of suprema of bounded sets is responsible for many constructions of real numbers. Most prominently, 2 is defined as sup{x Q x 2 < 2}.

5 12 Sequences Further, we pick up an example from section 1.1 now. Given a real number r > 0 and natural numbers n,m, we have a good intuition what r n m is: r n m is the m th root of r n. However, what shall we think of r 2? Or better: How is r 2 defined? We will answer this now with the aid of the axiom above. More generally, let r,p be real numbers. If r 1 we define r p := sup{r x x Q and x p}. We extend this definition for 0 < r < 1 by r p = ( 1 r ) p and show that We have r p = inf{r x x Q and x p} : r p = sup{( 1 r )x x Q and x p} = = sup{r x x Q and x p} = = sup{r x x Q and x p} = see below = inf{r x x Q and x p}. The last equality needs justification. Since r < 1 it is clear that for all x,y Q with x p y we have r x r y. Hence u := sup{r x x Q and x p} inf{r x x Q and x p} =: v. We now show that v u = 1 (and therefore v = u). We know already 1 v u and we show that for every real number δ > 0 we have v u < 1+δ. Choose n N with 1+nδ > 1 r and choose rational numbers x 1 p x 2 with x 2 x 1 < 1 n. By definition of u and v we have v rx1 and u r x2. Therefore Finally ( 1 r ) 1 n < 1+δ since 1 r v u rx1 r x2 = (1 r )x2 x1 by choice of n < 1+nδ as 1 r >1 and x2 x1< 1 n < ( 1 r ) 1 n. from the Binomial Theorem (1+δ) n Definition. Let (a n ) n N be a sequence. (a) (a n ) n N is called strictly increasing if a 1 < a 2 < a 3 <... and strictly decreasing if a 1 > a 2 > a 3 >... (a n ) n N is called strictly monotone if it is strictly increasing or strictly decreasing. (b) (a n ) n N is called increasing if a 1 a 2 a 3... and decreasing if a 1 a 2 a 3... (a n ) n N is called monotone if it is increasing or decreasing. The next theorem is a powerful tool to compute limits. We ll see examples shortly.

6 Bounded and monotone sequences Monotone convergence theorem. Every monotone and bounded sequence has a limit. Proof. Let (a n ) n N be our monotone sequence. We first assume that (a n ) n N is increasing, i.e. a 1 a 2 a 3... By assumption, (a n ) n N is bounded from above, which means that the value set A = {a n n N} is bounded from above. By the completeness axiom of R, A has a supremum s and we claim that s is the limit of (a n ) n N. To see this, take ε > 0. Since s is the least upper bound of A, s ε is not an upper bound of A, i.e. there is some N N with s ε < a N. Now if n N, then as (a n) n N is increasing as s=supa s ε < a N a n s. In particular a n s ε for all n N as required. It remains to do the case when (a n ) n N is decreasing. However, in this case the sequence ( a n ) n N is increasing (and again bounded). By what we have shown ( a n ) n N has a limit and by the scalar multiplication rule, also (a n ) n N has a limit.