# sole8 - n X k =1 | a k b k | ≤ n X k =1 a 2 k 1 2 n X k...

This preview shows pages 1–2. Sign up to view the full content.

ANALYSIS EXERCISE 8–SOLUTIONS 1. First, using the inequality 1 2 ( a + b ) ab , which is valid for all non-negative a and b we see that a n p for all n 2. So the sequence is bounded below. Next, we prove that it is decreasing. Indeed, a n +1 - a n = 1 2 p a n - a n = p - a 2 n 2 a n 0 . This implies that the sequence converges. Let l = lim n →∞ a n . Passing to the limit in the recurrence relation we obtain the equality l = 1 2 l + p l , which leeds to l 2 = p or l = p . 2. (a) s n = n X k =1 1 (3 k - 1)(3 k + 2) = 1 3 n X k =1 1 3 k - 1 - 1 3 k + 2 = 1 3 1 2 - 1 3 n + 2 . Therefore lim n →∞ s n = 1 6 . (b) s n = n + 2 - n + 1 + 1 - 2 = 1 n + 2 + n + 1 + 1 - 2. Therefore lim n →∞ s n = 1 - 2. 3. (a) Diverges since lim n →∞ n 2 n - 1 = 1 2 6 = 0. (b) Divereges since 1 n 1 n . (c) Diverges since lim n →∞ ( - 1) n n n +1 6 = 0. (d) Converges since 1 n n +1 1 n 3 / 2 . (e) Converges since 3 n n 5 n ( 3 5 ) n . 4. (a) From the convergence of X n =1 a n it follows that lim n →∞ a n = 0. Therefore ( N N )( n N ) [( n > N ) ( a n < 1)] . Hence for these n ’s a 2 n < a n . Apply the comparison test. (b) The assertion follows from the definition of the convergence of series and from the inequality

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n X k =1 | a k b k | ≤ n X k =1 a 2 k ! 1 / 2 n X k =1 b 2 k ! 1 / 2 . 1 2 ANALYSIS EXERCISE 8–SOLUTIONS There are several other proofs as well. For example, the result also follows from the simple inequality | a k b k | ≤ (1 / 2) a 2 k + (1 / 2) b 2 k . The sum of the right-hand side converges as it is the sum of con-vergent series. The left-hand side then converges by comparison. 5. (a) a n = 100 n n ! . Therefore lim n →∞ n √ a n = lim n →∞ 100 n √ n ! = 0 . By the Cauchy test the series converges. (b) a n = ( n √ n-1) n . Therefore lim n →∞ n √ a n = lim n →∞ ( n √ n-1) = 0 . By the Cauchy test the series converges. (c) a n = n 2 n . a n +1 a n = 1 2 n + 1 n . Hence lim n →∞ a n +1 a n = 1 2 . By the d’Alembert test the series converges....
View Full Document

{[ 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