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ratio-test

# ratio-test - a N 1 ∑ ∞ m =2 a N m Case 2 l> 1 This...

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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 specifically, | 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 + · · ·
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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...
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