# quiz4a - (a) ∞ X n =1 3 n + 2 n 6 n ∞ X n =1 3 n + 2 n...

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KEY QUIZ 4 Please show all your work. The answers without the solution will not be graded. 1. Prove the following: If the series X n =1 a n is convergent, then lim n →∞ a n = 0. PROOF: Suppose that the series converges to the limit L , i.e. X n =1 a n = L . Let s n = a 1 + a 2 + ··· + a n be the partial sum. Hence a n = s n - s n - 1 . Thus, lim n →∞ a n = lim n →∞ s n - s n - 1 = L - L = 0. ± 2. Complete the deﬁnition: A sequence { a n } has the limit L and we write lim n →∞ a n = L if for every ± > 0 ... there exists N > 0 such that n > N implies | a n - L | < ± . 3. Determine whether the series is convergent or divergent. If it is convergent, ﬁnd its sum.
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Unformatted text preview: (a) ∞ X n =1 3 n + 2 n 6 n ∞ X n =1 3 n + 2 n 6 n = ∞ X n =1 3 n 6 n + 2 n 6 n ! = ∞ X n =1 3 n 6 n + ∞ X n =1 2 n 6 n = ∞ X n =1 1 2 ! n + ∞ X n =1 1 3 ! n Since 1 2 < 1 and 1 3 < 1 the geometric series test tells us that the series converges. ∞ X n =1 1 2 ! 1 2 ! n-1 + ∞ X n =1 1 3 ! 1 3 ! n-1 = 1 / 2 1-1 / 2 + 1 / 3 1-1 / 3 = 1 + 1 2 = 3 2 (b) ∞ X n =1 1 5 n ∞ X n =1 1 5 n = 1 5 ∞ X n =1 1 n Since the Harmonic series diverges, this series also diverges....
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## This note was uploaded on 07/11/2011 for the course MAC 2312 taught by Professor Bonner during the Spring '08 term at University of Florida.

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