Chapter 6 SEQUENCES. sn 1+1,s 1 =1 is again an abbreviation for the sequence { 1, 1 2, 1 } n N ε,d (s n,s) <ε.

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1 Chapter 6 SEQUENCES 6.1 Sequences A sequence is a function whose domain is the set of natural numbers N. If s is a sequence, we usually denote its value at n by s n instead of s (n). We may refer to the sequence s as {sn} or by listing the elements {s1, s2,...}. This notation is somewhat ambiguous though, since a sequence is an ordered set, and should not be confused with the range of s = {sn : n N}. Some authors prefer the use of parentheses instead of braces, to emphasize this particular feature of sequences. Our notation will use braces for both. We call s n the n th term of the sequence and we often describe a sequence by giving a formula for the nth term. Thus {1/n} is an abbreviation for the sequence { } 1 1 1, 2, 3,.... Sometimes we may wish to change the domain of a sequence from N to N {0}, or to {n N : n m}. In this case we write {s n } n=0 or {s n} n=m respectively. If no mention is made to the contrary, we assume that the domain is just N. Sequences are sometimes also expressed in what we call recursive form. This means that a generic term sn is not expressed explicitly as a function of n, but instead as a function of some subset of previous terms {sn 1,s n 2..., s n k}. { Thus } s n : s n = s n 1 sn 1+1,s 1 =1 is again an abbreviation for the sequence { 1, 1 2, 1 } 3... Note that whenever the recursive definition for sn depends on k previous terms, for the sequence to be well defined it is necessary to provide the first k terms. 6.2 Convergence Convergent sequences Intuitively, we say that a given sequence {s n } converges to a certain value s whenever the sn become arbitrarily close to s for n arbitrarily large. It is clear from the above that we need a metric space to define what close means. In this section we will restrict to the metric space (R, ). The formal definition of convergent sequence in an arbitrary metric space is the following: Definition 224 A sequence {s n } in a metric space (X, d) is said to converge if there exists s X such that for every ε>0 there exists a real number N ε such that for all n N and n N ε,d (s n,s) <ε.

2 76 Sequences In this case we say that {sn} converges to s, or that s is the limit of the sequence {sn}, and write sn s, lim sn = s, or lim n s n = s. If a sequence does not converge, it is said to diverge. Remark 225 Notice that for a sequence to be convergent the definition requires its limit to belong to X, the set in which the sequence is defined. Take for example the sequence {1/n}. Even though it converges { to 0 in X = R, it fails } to do so in X = R ++. In the same fashion, the sequence x n : x n+1 = x2 n +2,x 2xn 1 =1 converges to 2 in R,but it fails to converge in Q. Example 226 The sequence {s n } with s n = 1 for all n N converges to 1 in R. The proof is trivial since d(sn, 1)=0<εfor all s n and for all ε>0. Example 227 Let us show that lim 1/n = 0 in (R, ). According to the definition, we have to show that for all ε > 0, M ε R, s.t. n N, n M ε, 1/n 0 = 1/n<ε.So take an arbitrary ε >0, ε R. From the Archimedean property of R there exits M ε N R such that M ε > 1/ε. Then it follows that for all n M ε we have 1/n 1/M ε <ε, which concludes the proof. Example 228 To prove that the sequence {1 +( 1) n } is divergent, let us suppose that {s n } converges to some real number s. Letting ε =1in the definition of convergence, we find that there exists a number M ε such that 1 +( 1) n s < 1. If n>m ε and n is odd, then 1 +( 1) n s = s < 1, so that 1 <s<1. On the other hand, if n>m ε and n is even, then we have 1 +( 1) n s = 2 s < 1, so that 1 <s<3. Since s cannot satisfy both inequalities, we have reached a contradiction. Thus the sequence {s n } must be divergent. So far we have seen examples where algebraic manipulation was enough to find an Mε and establish whether the definition was satisfied or not. However, for more complicated cases such a simple technique may not work. Example 229 Take for example the case of sn = n2 +2n n 3 5. lim s n =0. From the definition, given any ε>0, we want to make restricting n>3,we can remove the absolute value since n2 +2n n 3 5 We want to show that n2 +2n n 3 5 <ε.by > 0. Thus we want to know how big n has to be in order to make n2 +2n n 3 <ε.solving for this inequality would 5 be very messy, so we try to find some estimate of how large the left hand side can be. To do this we seek an upper bound for the numerator and a lower bound for the denominator. On the one hand we have n 2 +2n<an 2 for some a>1 if n> 2 (a 1). On the other we have n 3 5 > n3 pair a, b > 1 and restricting n>max b n 2 +2n n 3 5 for some b>1 if n> 5 3 b. Taking an arbitrary b 1 { } 3, we have that = n 2 (a 1), 2 +2n n b b 1 an 2 n 3 b = ab. n

3 Convergence 77 The right hand side can be easily made less than any ε. Just pick n ab. So ε now we are ready to organize this into a formal proof. Given ε > 0, take M ε = max { 3, 2 (a 1), 3 5 b, ab b 1 ε }. Then for all n N and n>m ε we have n 2 +2n n = n 2 +2n n 3 5 an 2 n 3 b = ab n <ε Hence lim n2 +2n n 3 5 =0. The use of upper bounds can be generalized by the following theorem Exercise 230 Suppose that lim s n = 0. If {tn} is a bounded sequence, prove that lim (sn t n)=0. Use this result to prove that lim sin(n) =0. n Theorem 231 Let {s n } and {a n } be sequences of real numbers and let s R. If for some k>0 and some m N we have s n s k a n, for all n>m and if lim a n =0, then it follows that lim s n = s. Proof. Given any ε>0, since lim an =0there exists N 1 R such that n N and n>n 1 implies that an <ε/k.now let N = max {m, N 1 }. Then for n>n we have n>mand n>n 1, so that s n s k a n <k ε k = ε Thus lim sn = s. Another useful theorem to show convergence of sequences by use of bounding sequences is the following. Theorem 232 ( Sandwich theorem) Suppose xn yn zn M is fixed, lim n x n = l, and lim n z n = l, then lim n y n = l. for n>m, where Exercise 233 Prove the Sandwich theorem. There is also a necessary condition for convergence that is very useful both in proofs of convergence and in proving that a sequence does not converge. It states that for n large enough, any two terms in a sequence must be arbitrarily close to each other.

4 78 Sequences Theorem 234 If a sequence {s n } converges, then for all ε>0 there is N R such that for all n 1,n 2 N, n 1,n 2 >N, sn 1 s n2 <ε. Proof. Since {sn} converges, for any ε>0 there is N R such that for all n N,n > N we have s n s < ε 2. In particular, if we take n 1 >N and n 2 >N we have by the triangle inequality sn 1 s n2 s n1 s + s s n2 < ε 2 + ε 2 = ε. Exercise 235 Prove that the reciprocal may not be true. That is, prove that you can have a sequence whose terms are arbitrarily close for n sufficiently large, but still the sequence does not converge Properties of Convergent Sequences In this section we derive two important properties of convergent sequences. But first we need a definition. Definition 236 Asequence {s n } is bounded if its range {s n : n N}is a bounded set, that is, if there exists M 0 such that sn M for all n N. Theorem 237 (Boundedness) Every convergent sequence is bounded. Proof. Let {s n } be a convergent sequence and let s =limsn. From the definition of convergence with ε =1, we obtain N R such that s n s < 1 whenever n > N. Thus for n > N the triangle inequality implies that sn < s +1. Let M = max{ s 1,..., s N, s +1}, then we have sn M for all n N, so {sn} is bounded. It is easy to see that the converse of this theorem is not true. We saw in a previous example the sequence {1+( 1) n }, which is bounded but diverges. Theorem 238 (Uniqueness) If a sequence {sn} converges, then its limit is unique. Proof. Let {sn} be a sequence and suppose that it converges both to x and y, with x y. Let x y = d > 0 be the distance between x and y. Convergence of {sn} to x implies that for any ε>0, there exists N 1 R such that s n x <ε, for every n>n 1. In particular this holds for ε = d 3. Similarly, convergence of {s n} to y implies that for any ε>0, there exists N 2 R such that s n y <ε, for every n>n 2, and in particular this holds for ε = d 3. Therefore if n>max {N 1,N 2 } we have 0 <d= x y = x sn + s n y x s n + s n y < 2 d 3 But this implies d < 2 d, which is a contradiction given that d >0. So it must be 3 d =0, and thus x = y.

5 Convergence Convergence and topological properties of sets In the previous chapter, we saw that a set X was closed if and only if it contained all its accumulation points. Here we will present the same result based on the theory of sequences. But in order to do that, we have first to prove two other theorems. The first one states that if xn x, then x is an accumulation point for the set described by the range of the sequence, and hence every neighborhood of x contains infinitely many points of {xn}. The second theorem establishes that it is equivalent to say that a point belongs to the closure of a given set E and to say that such a point is the limit of a sequence completely contained in E. Theorem 239 {xn} X converges to x X if and only if every neighborhood of x contains {xn} for all but finitely many n. Proof. If {xn} converges to x then for each ε > 0 there exists N R such that for all n N, n > Nd (xn,x) <ε. This means that any neighborhood of x of radius ε has at least one (hence infinitely many) elements of {xn}. Hence there must be at most only finitely many elements of {xn} outside any neighborhood of x. Suppose now that every neighborhood of x contains {xn} for all but finitely many n. Hence for each neighborhood N (x, ε), N R such that for all n N, n>n,xn N (x, ε). This suffices to conclude that {xn} converges to x. Theorem 240 Let E X. Then x E if and only if there exists a sequence {xn} in E such that x =lim n x n. Proof. Suppose there exists a sequence {x n } E that is convergent to x X but x/ Ē. Therefore x ĒC. Since ĒC is open, there exists a neighborhood N (x, ε) Ē C such that N (x, ε) {x n : n N} =. Therefore, for such ε there is no N R such that for all n>n xn x <ε,contradicting the assumption that x =lim n x n. Then x Ē. Suppose now that x Ē. Then x E or x bd E. This implies that for every ε>0, N (x, ε) E. We can therefore construct a sequence {xn} in E, being ( xn any point such that xn N x, 1 ) n E. By the Archimedean property of the reals, for all ε>0 there exists N N such that 1/N < ε, and by construction of the sequence there exists x N {x n } such that x N x < 1/N < ε. Therefore x =lim n x n. We have therefore seen that any limit point of E can be expressed as a limit of a convergent sequence in E. So, this theorem tells us that any nonempty closed set contains convergent sequences. Moreover, we know that if in a given set E every convergent sequence has a limit in the set, then the set is closed. Therefore we can define the closedness of a set in a metric space using convergent sequences. Theorem 241 A set E X is closed if and only if every convergent sequence in X completely contained in E has its limit in E. Proof. Let E be a closed set and let {xn} be a sequence in E converging to x X. By the previous theorem we know that x Ē. Since E is closed, then E = Ē and therefore x E.

6 80 Sequences Suppose now that every convergent sequence in X completely contained in E has its limit in E, but E is not closed. Then there exists x E Ē and x / E. As in the previous theorem there exists a sequence {x n } E such that x n x Ē E. But this contradicts that every convergent sequence {x n } E must have its limit in E. Therefore E is closed. 6.3 Limit and comparison theorems In the previous section, we saw that the definition of convergence may sometimes be messy to use even for sequences given by relatively simple formulas. In this section we will derive some basic results that will greatly simplify our work. The first theorem is a very important result showing that algebraic operations are compatible with taking limits. Theorem 242 Consider two sequences {x n } and {yn} such that xn x and yn y. Then the following properties hold: 1) xn + yn x + y; 2) xnyn xy; 3) If xn 0, n and x 0, then yn/xn y/x. Proof. 1) Using the triangular inequality, we know that x n + y n x y x n x + y n y. Now given ε > 0, since xn x there exists N 1 R such that n > N 1 implies x n x < ε 2. Similarly there exists such N 2 R such that n>n 2 implies y n y < ε 2. Thus, if we let N =max{n 1,N 2 }, then n>n implies that (xn + y n ) (x + y) x n x + y n y <ε. Therefore we conclude that lim (xn + yn) =x + y. 2) Observe that xnyn xy = xn(yn y)+y(xn x). Since {xn} is convergent then it is bounded. Let M 1 = sup x n and M =max {M 1, y }. Hence xnyn xy M 1 yn y + y x n x M ( y n y + x n x ). Now given ε>0 there exist N x and N y such that x n x < ε 2M and y n y < ε for all n>n x and n>n y respectively. Let N =max{n x,n y }. Then n>n implies that x n y n xy M ( yn y + xn x ) < M ε 2M + M ε 2M = ε. Thus lim x n y n = xy. 3) Since y n /x n = y n (1/x n ), it suffices from (2) to show that lim (1/x n )=1/x. That is, given ε>0, we must make 1 x n 1 x = x x n x n x <ε 2M

7 Limit and comparison theorems 81 for n sufficiently large. To get a lower bound on the denominator, we note that since x 0, there exists N 1 such that n>n 1 implies x n x < x 2. Thus for n>n 1 we have x xn < x 2 x x n < x 2 x 2 < x n and x x n x n x x n x < 2 x n x = x n x x 2 But since x n x, for every ε>0 there exists also N 2 such that for n>n 2 we have x n x < x 2 ε 2. Let N =max{n 1,N 2 }. Then n>n implies x x n x n x < 2 x 2 x n x <ε Hence lim 1/xn =1/x. A particular case of (1) is a + xn a + x, a R when xn x, and of (3) is axn ax, a R when xn x. Another useful fact is that the order relation is preserved when taking limits. The proof is left as an exercise. Theorem 243 Suppose that {yn} and {zn} are convergent sequences with lim yn = s and lim zn = z. If yn zn for all n>m, for certain M N, then y z. Exercise 244 Prove the previous theorem Infinite Limits The sequence sn = n is clearly not convergent, since it is not bounded. But its behavior is not the least erratic: the terms get larger and larger. Although there is no real number that the terms approach, we would like to say that sn goes to. We make this precise in the following definition. Definition 245 A sequence {x n } of R diverges to + if M > 0, N > 0 such that n > N, xn > M. We write xn + or lim n x n =+. A sequence {xn} diverges to if M > 0, N > 0 such that n > N, xn < M. We write xn or lim n xn =. The technique of developing proofs for infinite limits is similar to that for finite limits, and is closely related to theorem243. We leave the proof as an exercise.

8 82 Sequences Theorem 246 Suppose that {sn} and {tn} are sequences such that sn tn for all n>m, for a certain M N. (a) If lim sn = +, then lim tn = +. (b) If lim tn =, then lim sn =. Exercise 247 Prove the previous theorem. Exercise 248 Prove the following 1) lim ( n +1 n ) =0 2) lim n n 5 +3 =0 3) lim n! n n =0 6.4 Monotone sequences and Cauchy sequences In the preceding two sections, we have seen a number of results that enable us to show that a sequence converges. Unfortunately, most of these techniques depend on our knowing what the limit of the sequence is before we begin. Often in applications it is desirable to be able to show that a given sequence is convergent without knowing precisely the value of the limit. In this section we obtain two important theorems that do just that Monotone Sequences Definition 249 A sequence {x n } of real number is increasing if x n+1 x n (strictly increasing if x n+1 >x n ) n N; decreasing if x n+1 x n (strictly decreasing if x n+1 <x n ), n N and monotone if it is increasing or decreasing (strictly monotone if it is strictly increasing or strictly decreasing). Example 250 The sequence of real numbers {1/n} is decreasing (thus monotone) and bounded. The constant sequence x1 = x2 =... = 1is increasing, decreasing, and bounded. Both of them converge. Example 251 The sequence {( 1) n } is bounded, but it is not monotone. The sequence {2 n } is monotone but it is not bounded. None of them converge. As you may have imagined, we can generalize these examples in a theorem Theorem 252 (Monotone Convergence Theorem) A monotone sequence of real numbers {x n } converges if and only if it is bounded. Proof. We already know that a convergent sequence is bounded. Hence we just need to show that a monotone and bounded sequence of real numbers is convergent. Suppose {x n } is a bounded increasing sequence. Let S denote the nonempty bounded set {x n : n N}. By the completeness axiom S has a least upper bound, and we let x = sup S. We claim that lim xn = x. Given any ε>0, x ε is not an upper bound

9 Monotone sequences and Cauchy sequences 83 for S. Thus there exists an integer N such that x N >x ε. Furthermore, since {x n } is increasing and x is an upper bound for S we have x ε<x N x n x<x+ ε or equivalently x n x <ε for all n N. Hence lim x n = n. In the case when the sequence is decreasing, let x =inf S and proceed in a similar manner. Exercise 253 Consider the sequence { x n : x n+1 = x2 n +2 2xn,x 1 =1 2,{xn} is decreasing and bounded below. In particular, show that x n n 2 (try induction). Thus we know that {x n } is convergent in R. that {x n } converges to 2(Hint: use theorem 234). }. Show that for n 2 < x n+1 < Show also Cauchy sequences When a sequence is convergent, we saw in theorem 234 that the terms get close to each other as n gets large. It turns out that this property ( called the Cauchy property) in some cases is actually sufficient to imply convergence. Definition 254 A sequence {xn} in a metric space (X, d) is said to be a Cauchy sequence if for all ε>0, there exists a number M ε s.t. for all n, m N, n,m M ε d(x n,xm) <ε. Example 255 The sequence {1/n} R ++ is Cauchy. For each ε>0 take M ε = 2 ε, implying that for all n, m N and n,m>m ε, 1/n 1/m < 1/n + 1/m < 2 1 < Mε ε. Then the sequence is Cauchy. However, it is not convergent since 0 does not belong to R ++. Exercise 256 Prove that the sequence {( n 1/n)} is not Cauchy. i=1 As we saw in theorem 234, convergence is a sufficient condition for a sequence to be a Cauchy sequence. Theorem 257 Every convergent sequence is a Cauchy sequence. Proof. See theorem 234. Exercise 258 Show that every Cauchy sequence is bounded. Its contrapositive If a sequence is not Cauchy, then it is not convergent is very a useful tool to prove that a sequence does not converge. Its converse however, is not true as we have shown in the previous example. However, there are some metric spaces where Cauchy sequences do converge. These spaces are called complete.

10 84 Sequences Definition 259 A metric space (X, d) is called complete if every Cauchy sequence in X converges in X. Theorem 260 The metric space (R,d) is complete. Proof. Suppose that {s n } R is a Cauchy sequence and let S = {s n : n N} be the range of the sequence. We consider two cases, depending on whether S is finite or infinite. If S is finite, then the minimum distance ε between distinct points of S is positive. Since {sn} is Cauchy, there exists a number N such that m,n>n implies that sn sm <ε.let n 0 be the smallest integer greater than N. Given any m>n, sm and sn 0 are both in S, so if the distance between them is less than ε, it must be zero (since ε is the minimum distance between distinct points in S). Thus s m = s n0 for all n>n.it follows that lim s n = s n0. Now suppose that S is infinite. From the previous exercise we know that S is bounded. Thus from the Bolzano Wierestrass theorem, there exists a point s in R that is an accumulation point of S. We claim that {s n } converges to s. Given any ε > 0, there exists N R, such that s n s m <ε whenever n,m>n. Since s is an accumulation point of S, the neighborhood N ( s, ε ) = ( s ε,s+ ) ε contains infinitely many points of S. Thus in particular there exists an integer m>n such that s m N ( s, ε ). Hence, for any n>n we have 2 sn s = sn sm + sm s sn sm + sm s < ε 2 + ε 2 = ε and lim sn = s. Therefore every Cauchy sequence in R is a convergent sequence, which implies that (R,d) is a complete space. Corollary 261 Every Cauchy sequence in R is convergent. Note that in the proofs of both theorem, the completeness axiom of R is the key factor. So, you can naturally guess that these theorems do not hold for a sequence of Q, as it is the case. 6.5 Subsequences Definition 262 Let {s n } n=1 be a sequence and let {n k} k=1 be any sequence of natural numbers such that n 1 <n 2 <n 3 <...The sequence {x nk } k=1 quence of {x n } n=1. is called a subse- When viewed as a function, a subsequence of a sequence s : N R is the composition of s and a strictly increasing function n : N N. That is, s n : N R is a subsequence of the sequence s : N R. Since this functional notation is cumbersome, we prefer to use subscripts and write sn k instead of s (n (k)). It is important to note that the index of the subsequence is k, not nk and not n.

11 Subsequences 85 If we delete a finite number of terms in a sequence and renumber the remaining ones in the same order, we obtain a subsequence. In fact, we may delete infinitely many of them as long as there are infinitely many terms left. It is possible to do so because a countable set has countable subsets (theorem 157). Example 263 Given the sequence {x1,x 2,x 3,...,x n,...} the following represents a subsequence {x1, x3, x5,...,x2n+1,...} and it is obtained by taking into account only the odd terms of the original sequence. Exercise 264 Let {n k } k=1 be a sequence of natural numbers such that n k <n k+1 for all k N. Use induction to show that nk k for all k N. Theorem 265 A sequence {s n } converges to a real number s if, and only if, every subsequence of {s n } also converges to s. Proof. Let {s n } converge to s and take any subsequence {s nk }. We know that for each ε> 0 there exists a number N such that n>n implies d(s n,s) <ε. Thus when k > N,we apply the result of the previous exercise to obtain that n k k>n and therefore d(s nk, x) <ε. Suppose now that every subsequence {s nk } converges to s. Hence {s n } converges as well because it is also a subsequence of itself. Example 266 One application of this theorem is in finding the limit of a convergent sequence. Suppose that 0 <x<1 and consider the sequence defined by s n = x 1/n. Since 0 <x 1/n < 1 for all n, {s n } is bounded. Since x 1/n+1 x 1/n = x 1/n+1 ( 1 x n+1 n then {sn} is an increasing sequence. ) > 0, for all n Thus by the monotone convergence theorem {sn} converges to some number, say s. Now, for each n, s 2n = x 1/(2n) = s n. Since {s 2n} is a subsequence of {sn} then it also converges to s. Moreover, we have that lim sn = lim sn, and this implies s =lim sn =lim s2n =lim s n = lim sn = s. It follows that s 2 = s, so that s =0or s =1. Since s 1 = x>0 and s n is increasing, so s 0. Hence lim x 1/n =1 Exercise 267 Prove that if {sn} is a sequence of nonnegative terms and sn s, then lim sn = lim sn. Example 268 Another application of the previous theorem is in showing that a sequence is divergent. For example, if the sequence s n n =( 1) were convergent to some number s, then every subsequence would also converge to s. But {s 2n} converges to 1 and {s 2n+1 } converges to -1. We conclude that {s n } is not convergent.

12 86 Sequences So far we know that every subsequence of a convergent sequence converges to the same limit and that in a divergent sequence there exists a subsequence that diverges. Now the question is whether we can find a convergent subsequence in a divergent sequence. A version of the Bolzano-Wierestrass theorem for sequences says that if the sequence is bounded this is so. Theorem 269 Every bounded sequence of real numbers has a convergent subsequence. Proof. The proof is essentially the same as the Bolzano-Weierstrass theorem we have seen in the previous chapter and is left as an exercise. While an unbounded sequence may not have any convergent subsequence, it will contain a subsequence that has an infinite limit. We leave the result as an exercise. Exercise 270 Show that every unbounded sequence contains a monotone subsequence that has either + or - as a limit Limit Superior and Limit Inferior Definition 271 Let {s n } be a bounded sequence. A subsequential limit s is any real number that is the limit of some subsequence of {sn}. If S is the set of all subsequential limits of {sn}, then we define the limit superior to be lim sup n s n = sup S. Similarly, we define the limit inferior of {s n } to be lim inf n s n =inf S. Note that in the definition we require {sn} to be bounded. Thus theorem 269 implies that {sn} contains a convergent subsequence, so that the set S of subsequential limits will be nonempty. It will also be bounded, since {sn} is bounded. The completeness axiom then implies that sup S and inf S both exist as real numbers. Example 272 Let s n =( 1) n +(1/n). Then lim sup n s n =1, lim inf n s n = 1. It should be clear that we always have lim inf n s n lim sup n s n. Now, if {s n } converges to some number s, then all its subsequences converge to s, so we have lim inf n s n = lim sup n s n. The converse is also true. lim inf n s n < lim sup n s n, then we say that {s n } oscillates. If it happens that

13 Subsequences 87 Theorem 273 For a real valued sequence {s n }, lim n s n = s if and only if lim sup n s n = lim inf n s n = s. Proof. If sn converges to s so do all its subsequences. Hence S is singleton and sup S = inf S. Suppose now that sup S =infs = s. This means that S is singleton so that every converging subsequence of {sn} converges to s. Hence {sn} converges to s. Theorem 274 Let xn,y n be bounded sequences and xn yn for n>m,where M is fixed. Then lim sup n x n lim sup n y n, and lim inf n x n lim inf n y n. Proof. Let X =limsup n x n and Y = lim sup n y n and suppose that X> Y. Hence for some {x nk } and {y nk } subsequences of {x n } and {y n } we have that for every ε>0 there exist an N ε,x and N ε,y such that for all n k > N = max{n ε,x,n ε,y }, x nk X < ε and yn k Y < ε. Take ε <(X Y )/2 and let n = max{n,m}. Hence, for all n k > n, we have that x nk yn k >X Y 2ε >0 which gives a contradiction, meaning that lim sup n x n lim sup n y n. The part relative to lim inf can be proved in a similar way. Exercise 275 Let s n = n sin 2 ( n π 2 superior and the limit inferior of {sn}. ). Find the set S of subsequential limits, the limit Unbounded sequences There are some occasions when we wish to generalize the notion of the limit superior and the limit inferior to apply to unbounded sequences. There are two cases to consider for the limit superior, with analogous definitions applying to the limit inferior. 1. Suppose that {sn} is unbounded above. Then exercise 270 implies that there exists a subsequence having + as its limit. This prompt us to define lim sup n s n = Suppose that {sn} is bounded above but not below. If some subsequence converges to a finite number, we define lim sup n s n to be the supremum of the set of subsequential limits. Essentially, this coincides with the original definition. If no subsequence converges to a finite number, then we must have lim sn =, so we define lim sup n s n =. Thus for anysequence {sn}, lim sup n s n always exists as either a real number, + or. When k R and α = lim sup n s n, then writing α>kmeans that α is a real number greater than k or that α =+. Similarly, α<kmeans that α is a real number less than k or that α =.

14 88 Sequences Some special sequences The following are very common sequences. You need to be able to prove the results stated. Exercise 276 Prove the following statements 1. If p>0 then lim n ( 1 n )p =0; 2. If p>0 then n lim n n p =1; 3. lim n n =1; 4. If p>0 and α R, then lim n nα =0; (1+p) n 5. If x < 1, then lim n xn = References S.R. Lay, Analysis with an Introduction to Proof.Chapter 4. Third Edition. Prentice Hall. A. Matozzi, Lecture Notes Econ 897 University of Pennsylvania Summer Rudin, Principles of Mathematical Analysis. Chapter 3.