5

Sequences

Let (X, d) be a metric space. A sequence in X is a function from N to X. If ϕ : N → X is a sequence we write (xn ), (xn )n∈N

or

(x0 , x1 , . . .)

for ϕ, where xn = ϕ(n) is the nth term of the sequence ϕ = (x0 , x1 , . . .). Sequences in R are called real sequences. Definition 5.1. A sequence (xn ) in X converges to x if for every neighborhood U of x there exists N ∈ N such that xn ∈ U

for all n ≥ N .

In this case we write xn → x

or

lim xn = x

n→∞

or

lim xn = x. n

A sequence (xn ) which does not converge is called divergent. Proposition 5.2. The following are equivalent. (i) limn xn = x. (ii) For every ε > 0, there is N ∈ N such that xn ∈ B(x, ε) for all n ≥ N . (iii) For every ε > 0, there is N ∈ N such that d(xn , x) < ε for all n ≥ N Proposition 5.3. (i) (Uniqueness of the limit) A sequence cannot have more than one limit. (ii) (Boundedness) If (xn ) converges to x, then there is r > 0 such that xn ∈ Br (x). Proof of (ii). Assume that (xn ) converges to x. By (iii) of the previous proposition with ε = 1, there is N so that xn ∈ B1 (x) for all n ≥ N . Now take r be any number bigger or equal to d(x1 , x), . . . , d(xN −1 , x) and 1. Then xn ∈ Br (x) for all n, as claimed.  Definition 5.4. Let ϕ = (xn ) be a sequence in X and let φ : N → N be a strictly increasing function. Then ϕ ◦ φ : N → X is called a subsequence of ϕ = (xn ). We write (xnk ) for the subsequece ϕ ◦ φ where φ(k) = nk . Since ψ is strictly increasing n1 < n2 < n3 < . . .. 20

Proposition 5.5. If (xn ) is a convergent sequence with limit x, then every subsequence (xnk ) of (xn ) converges to x. Proof. Let (xnk ) be a subsequence of (xn ). Take a neighborhood U of x. Since lim xn = x, there is N ∈ N such that xn ∈ U for all n ≥ N . From the definition of a subsequence, nk ≥ k for all k ∈ N. So, nk ≥ N for all k ≥ N . Thus xnk ∈ U for all k ≥ N . Hence (xnk ) converges to x as claimed.  Definition 5.6. A sequence (xn ) in a metric space (X, d) is said to be Cauchy if for every ε > 0, there exists N such that d(xn , xm ) ≤ ε for all n, m ≥ N. Proposition 5.7. Let (X, d) be a metric space. (i) Every Cauchy sequence is bounded. (ii) Every convergent sequence in X is Cauchy. Proof of (ii). Assume that (xn ) converges to x. Then, given ε > 0, there is N such that d(xn , x) < ε/2 for all n ≥ N.. Hence

d(xn , xm ) ≤ d(xn , x) + d(x, xm ) < ε/2 + ε/2 = ε

for all n, m ≥ N . So, (xn ) is a Cauchy sequence as claimed.

5.1



Real sequences

Proposition 5.8. Let (xn ) and (yn ) be convergent sequences in R and let α ∈ R. Then (i) The sequence (xn + yn ) converges and lim(xn + yn ) = lim xn + lim yn . (ii) The sequence (αxn ) converges and lim(αxn ) = α lim xn . Proof of (i). Take ε > 0. Since (xn ) and (yn ) converge to x and y, respectively, there are N1 and N2 such that |xn − x| < ε/2

for n ≥ N1

and

|yn − y| < ε/2

for n ≥ N2 .

Consequently, |(xn + yn ) − (x + y)| = |(xn − x) + (yn − y)| ≤ |xn − x| + |yn − y| ≤ ε/2 + ε/2 = ε

for all n ≥ N = max{N1 , N2 }. Hence the sequence (xn + yn ) converges and lim(xn + yn ) = lim xn + lim yn , as claimed.  21

Note that above proposition holds true if instead of real sequences one considers sequences in a normed space (X, k·k). We call a real sequence (xn ) a null sequence if it converges to 0. That is, for every ε > 0, there is N ∈ N such that |xn | < ε for all n ≥ N . Proposition 5.9. Let (xn ) and (yn ) be sequences in R. (i) If (xn ) is a null sequence and (yn ) is a bounded sequence, then the sequence (xn yn ) is a null sequence. (ii) If (xn ) and (yn ) are convergent sequences, then the sequence (xn yn ) converges and lim(xn yn ) = lim xn lim yn . (iii) If (xn ) converges to x and x 6= 0, then almost all terms of (xn ) are nonzero and the sequence (1/xn ) converges to 1/x. Proof of (iI). Suppose that lim xn = x and lim yn = y. By (ii) of Proposition 5.3, |xn | ≤ A and |yn | ≤ B for all n. Moreover, given ε > 0, there are N1 and N2 so that |xn − x| < ε/2(A+|y|) for n ≥ N1

and

|yn − y| < ε/2(A+|y|) for n ≥ N2 .

Consequently, |(xn yn ) − (xy)| = |xn (yn − y) + y(xn − x)| ≤ |xn | |yn − y| + |y| |xn − x| ε ε + |y| =ε ≤ A |yn − y| + |y| |xn − x| ≤ A 2(A + |y|) 2(A + |y|) for all n ≥ N = max{N1 , N2 }. Hence the sequence (xn yn ) converges and lim(xn yn ) = lim xn lim yn , as claimed.  Proposition 5.10. (i) Let (xn ) and (yn ) be convergent sequences in R and xn ≤ yn for infinitely many n. Then lim xn ≤ lim yn . (ii) Let (xn ) and (yn ) and (zn ) be sequences in R and xn ≤ yn ≤ zn for almost all n. If lim xn = lim zn = a, then (yn ) converges to a. Proof of (ii). Take ε > 0. Then here are N1 and N2 so that |xn − a| < ε for n ≥ N1

and

|zn − a| < ε

for n ≥ N2 .

Hence a − ε < xn < a + ε

for n ≥ N1

and a − ε < zn < a + ε for n ≥ N2 22

and since xn ≤ yn ≤ zn , a − ε < xn ≤ yn ≤ zn < a + ε for all n ≥ N = max{N1 , N2 }. This means that |yn − a| < ε for all n ≥ N.  Definition 5.11. A sequence (xn ) in R is increasing, if xn ≤ xn+1 for all n ∈ N, and decreasing if xn+1 ≤ xn for all n ∈ N. If (xn ) is either increasing or decreasing, then it is called monotone. Proposition 5.12. Every increasing (or decreasing) bounded sequence (xn ) in R converges,and xn ր sup{xn | n ∈ N}

( or xn ց inf{xn | n ∈ N}).

Proof. Suppose that (xn ) is increasing and Set a = sup{xn | n ∈ N} ∈ R. By the characterization of the supremum, (a) xn ≤ a for all n. (b) given ε, there is N such that a − ε < xN . Since (xn ) is increasing, a − ε < xN ≤ xn ≤ a < a + ε for all n ≥ N . Hence (xn ) converges to x as claimed.



Example 5.13. Let a > 0 and α > 0. Define a sequence (xn ) by setting x0 = α and

  a xn + . (1) xn √ We claim that the sequence (xn√ ) converges to a. To see this we shall show that (xn ) is bounded below by a and it is decreasing. Consider the function

one finds that f ′ (x) = 12 1 − xa2 < 0 f (x) = 12 x√+ xa for x > 0. Differentiating √ ′ for x√∈ (0, a) and f√ (x) > √ 0 for x ∈ ( a, √ ∞). Hence f takes its minimum at a = a and f (x) ≥ f ( a) = a. So, xn ≥ a for all n ≥ 1. Next     1 a a 1 x2 − a ≥0 xn − xn+1 = xn − xn + xn − = = n 2 xn 2 xn 2xn xn+1 =

1 2

since x2n ≥ a. This means that xn ≥ xn+1 for all n ≥ 1 so that √ (xn ) is decreasing. Then the above proposition implies that xn → x where x ≥ a > 0 Then also xn+1 → x. So, taking limits in (1), we get 1 a x= x+ 2 x √ and solving for x one finds x = a.

23

Proposition 5.14. Every sequence contains a monotone subsequence. Proof. Consider a sequence (xn ). We call the m-th term xm a peak if xm ≥ xn for all n ≥ m. There are two cases to consider. Case 1. There are infinitely many peaks. List them according the increasing subscripts: xn1 , xn2 , . . .. Then, since each of them is a peak, we have xn 1 ≥ xn 2 ≥ xn 3 ≥ . . . . Hence (xnk ) is a decreasing subsequence of (xn ). Case 2. There are finitely many peaks. List them according to the increasing xm1 , xm2 , . . . , xml . Then set n1 = ml + 1. The n1 terms xn1 is not a peak and so there is an index n2 > n1 such that xn1 < xn2 . The term xn2 is also not a peak and there is n3 > n2 such that xn2 < xn3 . Continuing this way we obtain a strictly increasing subsequence (xnk ) of (xn ).  5.1.1

Infinite limits

Definition 5.15. A sequence of real numbers (xn ) converges to ∞, in symbols xn → ∞ or lim xn = ∞, if for every M > 0 there exists N such that xn ≥ M

for all n ≥ N .

Similarly, (xn ) converges to −∞, in symbols xn → −∞ or lim xn = −∞ if for every M < 0 , there is there exists N such that xn ≤ M 5.1.2

for all n ≥ N .

Some Special Sequences

(1) Let a ∈ R. Then an → 0 if |a| < 1

an diverges Suppose that |a| < 1. Then |a| =

1 1+h

if |a| > 1

for some h > 0. By the binomial theorem,

(1 + h)n =

n   X n k h ≥ nh k k=0

so that n

0 ≤ |an | = |a| =

24

1 1 ≤ . (1 + h)n nh

Since 1/nh → 0, an → 0 by the comparison test. Assume that |a| > 1. Then |a| = 1 + h for some h and then |an | = |an | = (1 + h)n ≥ nh, showing that (an ) is unbounded. (2) Let k ∈ N and let a ∈ R be such that |a| > 1. Then nk = 0. n→∞ an lim

To see this write a = 1 + h where h > 0 and apply the binomial theorem,   n   X n m n n(n − 1) · · · (n − k) k+1 n n h . a = (1 + h) = h ≥ hk+1 ≥ (k + 1)! m k + 1 m=0 So, 0≤

nk (k + 1)!nk ≤ k+1 k a h n(n − 1) · · · (n − k)

and the claimed follows using the comparison since the right hand side converges to 0 as n → ∞. (3) For all a ∈ R, an = 0. lim n→∞ n! √ (4) limn→∞ n n = √1. √ To see this write n n = 1 + an with an ≥ 0 (since it is easy to see that n n ≥ 1. So, (1 + an )n = n. By the binomial theorem,       n n 2 n(n − 1) 2 n 2 an . n = (1 + an )n = 1 + an + an + . . . ≥ 1 + an = 1 + 2 1 2 2 From this we get 0 ≤ a2n ≤

2 , n

i.e., 0 ≤

p √ n n − 1 = an ≤ 2/n

√ which by comparison implies that n n → 1. √ n (5) For all a > 0, limn→∞ a = 1. If a = 1, then there is nothing to prove. If a > 1, then √ √ 1na≤ nn √ which implies that ( n a) converges to 1. So, assume that 0 < a < 1, then 1/a > 1 p p √ so that n 1/a → 1. Since n a = 1/ n 1/a, the claim follows.

25

5.1.3

The limit superior and limit inferior

Let (xn ) be a bounded sequence. For each n ∈ N, let yn = sup xm = sup{xn , xn+1 , . . .}. m≥n

Then {yn } is decreasing, and it is bounded since (xn ) is bounded. By Proposition 5.12, (yn ) converges. The limit is called upper limit of (xn ), and is denoted by lim sup xn . Similarly, let zn = inf m≥n xm = inf{xn , xn+1 , . . .}. Then (zn ) is increasing, and it is bounded since (xn ) is bounded. The limit of (zn ) is called lower limit of (xn ), and is denoted by lim inf xn . If (xn ) is not bounded from above, then its upper limit is equal to ∞ and if (xn ) is not bounded from below, then its lower limit is equal to −∞. Summarizing lim sup xn = lim xn = inf sup xk = lim sup xk n≥m k≥n

n→∞ k≥n

lim inf xn = lim xn = sup inf xk = lim inf xk . n≥m k≥n

n→∞ k≥n

The basic properties of the upper and the lower limits are listed in the following proposition: Proposition 5.16. If (xn ) and (yn ) are sequences of real numbers, then: (a) lim sup(−xn ) = − lim inf xn and lim inf(−xn ) = − lim sup xn . (c) lim sup(axn ) = a lim sup xn and lim inf(axn ) = a lim inf xn for any a > 0. (d) lim sup(xn +yn ) ≤ lim sup xn +lim sup yn and lim inf xn +lim inf yn ≤ lim inf(xn + yn ). (e) lim inf xn ≤ lim sup xn , with equality if and only if (xn ) converges. In this case lim sup xn = lim xn . Theorem 5.17 (Bolzano-Weierstrass Theorem). Let (xn ) be a bounded sequence in R. Then (xn ) has convergent subsequence. Proof. By Proposition 5.14, (xn ) has a monotone subsequence (xnk ). Since (xnk ) is bounded, Proposition 5.12 implies that is converges.  Theorem 5.18 (Cauchy’s criterion). A sequence (xn ) of real numbers converges if and only if (xn ) is a Cauchy sequence. Proof. We already know that a convergent sequence is Cauchy (Proposition 5.7). Hence we have to prove that a Cauchy sequence (xn ) converges to some x ∈ R. By Proposition 5.7, (xn ) is bounded so that by Theorem 5.17, it contains a convergent subsequence, say xnk → x. We claim that xn → x. To see this, take ε > 0. Then there is K such that |xnk − x| < ε/2 for all k ≥ K.

26

Moreover, since (xn ) is Cauchy, there is N such that |xn − xm | < ε/2 for all n, m ≥ N . Take k such that nk ≥ N and let n ≥ N . Then |xn − x| ≤ |xn − xnk | + |xnk − x| < ε/2 + ε/2 = ε. Hence (xn ) converges to x, as claimed.



Example 5.19. Define (xn ) by setting x1 = 1, x2 = 2, and xn =

1 (xn−1 + xn−2 ) . 2

We claim that the sequence converges to x = 5/3. To see this we shall show that (xn ) is Cauchy and that the subsequence (x2n+1 ) converges to x = 5/3. First note that 1 ≤ xn ≤ 2 for all n. Next |xn − xn+1 | =

1 . 2n−1

1 Indeed, for n = 1, |x1 − x2 | = 1 = 210 . Assuming that |xn − xn+1 | = 2n−1 , we show that |xn+1 − xn+2 | = 21n . We have 1 1 1 |xn+1 − xn+2 | = xn+1 − (xn+1 − xn ) = |xn − xn+1 | = n , 2 2 2

as claimed. Consequently, if m > n,

|xn − xm | ≤ |xn − xn+1 | + |xn+1 − xn+2 | + . . . + |xm−1 − xm | 1 1 1 ≤ n−1 + n + . . . + m−2 2 2  2  1 1 1 1 = n−1 1 + + . . . + m−n−1 < n−2 2 2 2 2 and since 1/2n−2 → 0, it follows that indeed (xn ) is Cauchy. Now to find the limit of x we first show that x2n−1 < x2n . This holds true for n = 1. Assuming that x2n−1 < x2n , we show that x2n+1 < x2n+2 . From the assumption if follows that x2n−1
(x2n−1 + x2n ) = x2n+1 , 2 2

as claimed. From this and |x2n+1 − xn+2 | = 1/22n , it follows that x2n+2 = x2n+1 +

27

1 . 22n

Now use induction to prove that x2n+1 = 1 +

1 1 1 + . . . + 2n−1 . + 2 23 2

For n = 1 this is clear. Assuming the above equality we show that x2n+3 = 1 . We have 1 + 21 + 213 + . . . + 22n+1   1 1 1 x2n+3 = (x2n+2 + x2n+1 ) = + 2x2n+1 2 2 22n 1 1 1 1 = 1 + + 3 + . . . + 2n−1 + 2n+1 , 2 2 2 2 as claimed. Now, note that 1 1 1 + 3 + . . . + 2n−1 2 2 2  1 1 1 1 1 + 2 + 4 + . . . + 2n−2 =1+ 2 2 2 2   1 1 1 1 1 + + 2 + . . . + n−1 =1+ 2 4 4 4   1 5 2 1− n → , =1+ 3 4 3

x2n+1 = 1 +

where we have used the formula

Pn

k=0

ak =

28

1−an+1 1−a

provided that a 6= 1.