Cauchy's convergence criterion 2.24. Prove Cauchy's convergence criterion as stated on Page 27. Necessity. Suppose the sequence (a,) converges to /. Then, given any e>0, we can find N such that I,-1l <«2 for all p > N and In, -1l N Then, for both p>N and q> N, we have Sufficiency. Suppose Iw, -, | N and any e>0. Then all the numbers u, Hy.. lie in a finite interval; i.., the set is bounded and infinite. Hence, by the Bolzano-Weierstrass theorem there is at least one limit point-say, a. If a is the only limit point, we have the desired proof and lim .a. Suppose there are two distinct limit points-say, a and b-and suppose b>a (see Figure 2.1). By defini- tion of limit points, we have Iu,-al <(b - ay3 for infinitely many values of p (1) I, -bl <(b - ay3 for infinitely many values of g Figure 2.1 (2) 36 CHAPTER 2 Sequences Then, since b-a (b-) + (,- ) + (M, - a), we have Ib-al -b-as lb-w,l • I, -,l+I,-al Using Equations (1) and (2) in (3), we see that ,-,| > ( - ay3 for infinitely many values of p and q, thus contradicting the hypothesis that 4,-,lce for p. q>Nand any e>0. Hence, there is only one limit point and the theorem is proved. (3)

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Cauchy's convergence criterion
2.24.
Prove Cauchy's convergence criterion as stated on Page 27.
Necessity. Suppose the sequence (u,} converges to l. Then, given any e>0, we can find N such that
lu,-1| <e2 for all p>N and |u, -1| <2 for all q> N
Then, for both p> N and q > N, we have
Sufficiency. Suppose lu,- u, |<e for all p, q > N and any e>0. Then all the numbers u, Hy . 1... lie in
a finite interval; i.ce., the set is bounded and infinite. Hence, by the Bolzano-Weierstrass theorem there is at
least one limit point-say, a.
If a is the only limit point, we have the desired proof and lim .=a.
Suppose there are two distinct limit points-say, a and b-and suppose b>a (see Figure 2.1). By defini-
tion of limit points, we have
I, - al <(b-av3 for infinitely many values of
(1)
|u,-b| <(b - ay3 for infinitely many values of g
Figure 2.1
(2)
36
CHAPTER 2 Sequences
Then, since b-a (b -) + (H, - M,) + (4, - a), we have
Ib-al = b-a s lb-u,l + lu, -u,l + Iu, -al
(3)
Using Equations (1) and (2) in (3), we see that u, -u, > (b - ay3 for infinitely many values of p and
q. thus contradicting the hypothesis that |u,-u, <e for p, q>Nand any e>0. Hence, there is only one limit
point and the theorem is proved.
Transcribed Image Text:Cauchy's convergence criterion 2.24. Prove Cauchy's convergence criterion as stated on Page 27. Necessity. Suppose the sequence (u,} converges to l. Then, given any e>0, we can find N such that lu,-1| <e2 for all p>N and |u, -1| <2 for all q> N Then, for both p> N and q > N, we have Sufficiency. Suppose lu,- u, |<e for all p, q > N and any e>0. Then all the numbers u, Hy . 1... lie in a finite interval; i.ce., the set is bounded and infinite. Hence, by the Bolzano-Weierstrass theorem there is at least one limit point-say, a. If a is the only limit point, we have the desired proof and lim .=a. Suppose there are two distinct limit points-say, a and b-and suppose b>a (see Figure 2.1). By defini- tion of limit points, we have I, - al <(b-av3 for infinitely many values of (1) |u,-b| <(b - ay3 for infinitely many values of g Figure 2.1 (2) 36 CHAPTER 2 Sequences Then, since b-a (b -) + (H, - M,) + (4, - a), we have Ib-al = b-a s lb-u,l + lu, -u,l + Iu, -al (3) Using Equations (1) and (2) in (3), we see that u, -u, > (b - ay3 for infinitely many values of p and q. thus contradicting the hypothesis that |u,-u, <e for p, q>Nand any e>0. Hence, there is only one limit point and the theorem is proved.
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