121616949-math.271

121616949-math.271 - series were positive in general we...

This preview shows page 1. Sign up to view the full content.

10.7 The Ratio and Root Tests 257 Because lim n →∞ a n +1 a n = 1 5 , then when n is big enough, say n N for some N , a n +1 a n < 1 2 and a n +1 < a n 2 . So a N +1 < a N / 2, a N +2 < a N +1 / 2 < a N / 4, a N +3 < a N +2 / 2 < a N +1 / 4 < a N / 8, and so on. The general form is a N + k < a N / 2 k . So if we look at the series a N + a N +1 + a N +2 + a N +3 + · · · + a N + k + · · · , its terms are less than or equal to the terms of the sequence a N + a N 2 + a N 4 + a N 8 + · · · + a N 2 k + · · · = X k =0 a N 2 k = 2 a N . So by the comparison test, X k =0 a N + k converges, and this means that X n =0 a n converges, since we’ve just added the fixed number a 0 + a 1 + · · · + a N - 1 . Under what circumstances could we do this? What was crucial was that the limit of a n +1 /a n , say L , was less than 1 so that we could pick a value r so that L < r < 1. The fact that L < r (1 / 5 < 1 / 2 in our example) means that we can compare the series a n to r n , and the fact that r < 1 guarantees that r n converges. That’s really all that is required to make the argument work. We also made use of the fact that the terms of the
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: series were positive; in general we simply consider the absolute values of the terms and we end up testing for absolute convergence. THEOREM 10.39 The Ratio Test Suppose that lim n →∞ | a n +1 /a n | = L . If L < 1 the series ∑ a n converges absolutely, if L > 1 the series diverges, and if L = 1 this test gives no information. Proof. The example above essentially proves the ﬁrst part of this, if we simply replace 1 / 5 by L and 1 / 2 by r . Suppose that L > 1, and pick r so that 1 < r < L . Then for n ≥ N , for some N , | a n +1 | | a n | > r and | a n +1 | > r | a n | ....
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