1549901369-Elements_of_Real_Analysis__Denlinger_

118 Chapter 2 • Sequences

Proof. Suppose { Xn} is a Cauchy sequence. Then, taking E = 1 in Defini-

tion 2.7.1, :lno EN 3 m,n ~no=> lxm - Xnl < l. Then

n 2: no ::::} lxn - Xnol < 1

Xno + l -

Xno --

Xno -1 -

=> -1 < Xn - Xn 0 < 1

::::} Xn 0 - 1 < Xn < Xn 0 + 1.

------T---~--------• • I

• 
  
  
  
  • • • •
  
  
  
  

=~===~i====~==~~=== •I

Figure 2.6

As in the proof of Theorem 2.2.10, we define

A= min{x1, X2, · · · , Xn 0 , Xn 0 - 1} and B = max{x1, X2, · · · , Xn 0 , Xn 0 + 1}.

Then \:/n EN, A:::; Xn:::; B. That is , {xn} is bounded. •

lff.WI Theorem 2.7.4 Every Cauchy sequence converges.

Proof. Suppose {xn} is a Cauchy sequence. By Theorem 2.7.3, {xn} is

bounded. Hence, by the Bolzano-Weierstrass Theorem, it has a convergent sub-

sequence, say {xnk}, with Xnk --+ L.

To prove: Xn--+ L. Let E > 0. Since { xn} is a Cauchy sequence, :3 n 1 EN 3

m , n ~ n1 => lxm - xnl <~-Since Xnk --+ L , :3n 2 EN 3 k 2: n2 => lxnk - LI<

€

2· Let no= max{n 1 ,n2}. Then

m ~no => m ~ n1 and nm ~ m 2: n1 and nm 2: m 2: n2

€ €

::::} lxm - Xnm I < 2 and lxnm - LI < 2

=> lxm --:--Xnm I + lxnm - LI < €

=> lxm - Xnm + Xnm - LI < € by the triangle inequality

=> lxm - LI < €.

Therefore, Xn --+ L. •

Theorem 2. 7.5 Suppose that { Xn} is a sequence for which there is a constant

c -

C such that\:/n EN, lxn+i -xn l <-. Then {xn} is a Cauchy sequence; hence

2n

it converges.