4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES

4.4

131

Monotone Sequences and Cauchy Sequences

4.4.1

Monotone Sequences

The techniques we have studied so far require we know the limit of a sequence in order to prove the sequence converges. However, it is not always possible to …nd the limit of a sequence by using the de…nition, or the limit rules. This happens when the formula de…ning the sequence is too complex to work with. It also happens with sequences de…ned recursively. Furthermore, it is often the case that it is more important to know if a sequence converges than what it converges to. In this section, we look at two ways to prove a sequence converges without knowing its limit. We begin by looking at sequences which are monotone and bounded. These terms were de…ned at the beginning of this chapter. You will recall that in order to show that a sequence is increasing, several methods can be used. 1. Direct approach, simply show that an+1 2. Equivalently, show that an+1 3. Equivalently, show that positive.

an

an+1 an

an for every n.

0 for every n.

1 for every n if both an and an+1 are

4. If an = f (n), one can show a sequence (an ) is increasing by showing that f is increasing that is by showing that f 0 (x) 0. 5. By induction. We now state and prove an important theorem about the convergence of increasing sequences. Theorem 344 An increasing sequence (an ) which is bounded above converges. Furthermore, lim an = sup fan g. n!1

Proof. We need to show that given > 0, there exists N such that n N =) jan Aj < where A is the limit. Let > 0 be given. Since (an ) is bounded above, by theorem 158, fan g has a supremum. Let A = sup fan g. Let > 0 be given. then, A < A. By theorem 151, there exists an element of fan g, call it aN , such that A < aN A. Since (an ) is increasing, we have an aN for every n N . Therefore, if n N , A

< an () 0

=) jan

A () A

an
0 such that m; n < N =) jxm xn j < . That it we would like to have jxn Since

xm j < () 1 n

1 n

1 < m

1 1 1 < + m n m

1 1 1 + < , the result will follow. This will happen if both < n m n 2 2 1 2 that is when n > and < that is when n > . So, we see that if N is an m 2 2 integer larger than then m; n > N =) jxm xn j < .

If we make

::: 1

k

k 1 . n+1

1 n

.

4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES

135

n X 1 Example 356 Consider (xn ) where xn = k2 k=1 Let > 0 be given. We want to show that there exists an integer N > 0 such that m; n < N =) jxm xn j < (without loss of generality, let us assume that n > m). That it we would like to have

jxn

xm j


an+1 > bn+1 > bn . (b) Deduce that both (an ) and (bn ) converge. (c) Show that lim an = lim bn . Gauss called the common value of these limits the arithmetic-geometric mean. 6. Prove theorem 357. 7. Answer the why? parts in the proof of theorem 358. 8. Finish proving theorem 361.

138

4.5

CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES

Subsequences

Discuss subsequences, the Bolzano-Weierstrass theorem, limit sup and inf.

4.5.1

Exercises