notes4 - CM221A ANALYSIS I NOTES ON WEEK 4 TWO CONVERGENCE...

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CM221A ANALYSIS I NOTES ON WEEK 4 TWO CONVERGENCE TESTS Ratio Test Theorem. Assume that a n 6 = 0 for all n , and that lim n →∞ | a n +1 /a n | exists and is equal to c . If c < 1 then the series n =1 a n is absolutely convergent. If c > 1 then the series diverges. (If c = 1 nothing can be said.) Proof. Assume that c < 1. Then there exists a positive number b such that c < b < 1. From the definition of the limit it follows that | a n +1 /a n | 6 b for all sufficiently large n or, in other words, | a n +1 | 6 b | a n | for all n > n 0 , where n 0 is some positive integer depending on b . Then we have | a n 0 +1 | 6 b | a n 0 | , | a n 0 +2 | 6 b 2 | a n 0 | , | a n 0 +3 | 6 b 3 | a n 0 | , ..., | a n 0 + j | 6 b j | a n 0 | , ..., which implies that | a n | 6 | a n 0 | b - n 0 b n for all n > n 0 . Since the series n =1 b n converges, by the Comparison Theorem the series n =1 a n is absolutely convergent. Assume now that c > 1. Then there exists a positive number b such that 1 < b < c . From the definition of the limit it follows that | a n +1 /a n | > b for all sufficiently large n or, in other words, | a n +1 | > b | a n | for all n > n 0 , where n 0 is some positive integer depending on b . Then we have | a n 0 +1 | > b | a n 0 | , | a n 0 +2 | > b 2 | a n 0 | , | a n 0 +3 | > b 3 | a n 0 | , ..., | a n 0 + j | > b j | a n 0 | , ..., which implies that a n 0 + k + as k → ∞ . In this case the series n =1 a n diverges because a n do not converge to 0 as n → ∞ . n th-root Test Theorem. Let lim sup | a n | 1 /n = c . If c < 1 then the series n =1 a n is absolutely convergent. If c > 1 then the series diverges. (If c = 1 nothing can be said.) Proof. Assume that c < 1. Then there exists a positive number b such that c < b < 1. From the definition of the upper limit it follows that | a n | 1 /n 6 b
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