Ratio-test - a N 1 ∑ ∞ m =2 a N m Case 2 l> 1 This time choose ²> 0 small enough so that q:= l-²> 1 Then there is some N such that if

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

View Full Document Right Arrow Icon
Proof of Proposition 6.17 (Ratio Test) Patrick De Leenheer February 14, 2005 Proposition 1. Let n =1 a n be a series with lim n →∞ | a n +1 | | a n | = l. (1) If l < 1 , then the series n =1 a n converges. If l > 1 , then the series diverges. Proof. Case 1 : l < 1. From (1) follows that if we choose ² > 0 small enough such that q := l + ² < 1, then there is some N , such that if n > N , | a n +1 | | a n | < q. More speci±cally, | a N +2 | < q | a N +1 | , | a N +3 | < q | a N +2 | < q 2 | a N +1 | , . . . , | a N + m | < q m - 1 | a N +1 | , m > 1 . Now, since 0 < q < 1, it follows (geometric series!) that X m =2 q m - 1 | a N +1 | = ( 1 1 - q - 1) | a N +1 | and then the comparison test yields that m =2 a N + m converges. This implies that the series n =1 a n converges too because n =1 a n = a 1 + a 2 + · · · +
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a N +1 + ∑ ∞ m =2 a N + m . Case 2 : l > 1. This time choose ² > 0 small enough so that q := l-² > 1. Then there is some N such that if n > N , | a n +1 | | a n | > q In particular, | a N +2 | > q | a N +1 | , | a N +3 | > q | a N +2 | > q 2 | a N +1 | , . . . , | a N + m | > q m-1 | a N +1 | , ∀ m > 1 . Since q > 1, it is clear that it is impossible that lim n →∞ a n = 0 (in fact, lim n →∞ | a n | = + ∞ !), which is necessary for convergence of the series ∑ ∞ n =1 a n (see Theorem 6 . 9). 1...
View Full Document

This note was uploaded on 09/12/2011 for the course MAP 4305 taught by Professor Deleenheer during the Summer '06 term at University of Florida.

Ask a homework question - tutors are online