ia2sp10h4s

# The second and third difficulty are frequently

• Notes
• 4

This preview shows pages 1–4. Sign up to view the full content.

sequence actually converges to the assumed limit. The second and third difficulty are frequently interchangeable. We just overcame the first difficulty; if the sequence converges to anything it has to be to a . Back to the proof. The sequence { a n } is a Cauchy sequence, thus is bounded. (recall that Cauchy sequences are bounded!). In other words, there is some real number C > 0 such that k a n k 1 C for n = 1 , 2 , 3 . . . . For every K N we have K X k =1 | a nk | ≤ X k =1 | a nk | = k a n k 1 C for n = 1 , 2 , 3 , . . . . Since (it being a finite sum!) lim n →∞ K X k =1 | a nk | = K X k =1 | lim n →∞ a nk | = K X k =1 | a k |

This preview has intentionally blurred sections. Sign up to view the full version.

we see that K X k =1 | a k | ≤ C for all K , hence also k a k 1 = X k =1 | a k | ≤ C, proving a 1 . Finally, we need to show that k a n - a k 1 0 as n → ∞ . The problem her is to control well the tails of the sequences. Let ² > 0 be given. Because a 1 , there is K such that X k = K +1 | a k | < ² 2 . By the Cauchy condition, there is N such that k a n - a m k 1 < ²/ 2 whenever n, m N . Thus, for every p 1, n, m N , K + p X k = K +1 | a nk - a mk | ≤ k a n - a m k 1 < ² 2 . Keeping p fixed for a moment we let m → ∞ (the sum being finite, the limit can be interchanged with the sum) to conclude that K + p X k = K +1 | a nk - a k | < ² 2 for all p N , n N , n N . This being true for all p 1 it is also true for the limit as p → ∞ ; in other words, X k = K +1 | a nk - a k | < ² 2 if n N . Now (again because the sum is finite) lim n →∞ K X k =1 | a nk - a k | = 0 , thus there is N 1 such that K X k =1 | a nk - a k | < ² 2 if n N 1 . Let n max( N, N 1 ). Then k a n - a k 1 = X k =1 | a nk - a k | = K X k =1 | a nk - a k | + X k = K +1 | a nk - a k | < ² 2 + ² 2 = e proving that { a n } converges to a in 1 . NOTES: 1. If you got only 5 points for this exercise it is because I assumed that everybody would have no difficulty proving everything but the completeness, thus I emphasized that part of the problem. A 5 means that, in my opinion, your proof of completeness has profound flaws. But see also Note 3. 2. Several of you used (essentially) the following result without justification: If lim n →∞ a nk = a k , then (1) X k =1 | a k | = lim n →∞ X k =1 | a nk | . 2
Exchanging infinite processes is a delicate operation; it has to be done with care. For example, suppose we take a nk = 1 n 2 k if k 6 = n , k k=n For a fixed k , once n > k the sequence is the same as 1 / (2 k n ) 0 as n → ∞ , thus a k = lim n →∞ a nk = 0 for all k .

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.
• Spring '11
• Speinklo
• Metric space, Limit of a sequence, Hilbert space, Cauchy sequence, Banach space

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern