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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)

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)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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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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