quiz4_sol - so we can split the series into two series.) X...

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Solutions QUIZ 4 March 5, 2010 1. Show 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 = lim n →∞ s n = L . Where s n = a 1 + a 2 + ··· + a n is 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 ε > 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. X n =1 1 + 2 n 6 n X n =1 1 + 2 n 6 n = X n =1 1 6 n + 2 n 6 n ! = X n =1 1 6 n + X n =1 2 n 6 n = X n =1 1 6 ! n + X n =1 1 3 ! n (Note: Since | 1 6 | < 1 and | 1 3 | < 1 the geometric series test tells us that the series converges
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Unformatted text preview: so we can split the series into two series.) X n =1 1 6 ! 1 6 ! n-1 + X n =1 1 3 ! 1 3 ! n-1 = 1 / 6 1-1 / 6 + 1 / 3 1-1 / 3 = 1 5 + 1 2 = 7 10 4. Determine whether the sequence is convergent or divergent. If it is convergent, nd the limit. 3 + 5 n 2 n + n 2 n =1 lim n 3 + 5 n 2 n + n 2 = lim n (3 + 5 n 2 )1 /n 2 ( n + n 2 )1 /n 2 = lim n 3 /n 2 + 5 1 /n + 1 = 0 + 5 0 + 1 = 5 Have a good Spring Break! Remember not to drink and derive (or integrate)....
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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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