But n 1 n n 1 while n n n 1 n 1 1 n 1 parenleftbigg n

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But ( n + 1)! n ! = n + 1 , while n n ( n + 1) n +1 = 1 n + 1 parenleftbigg n n + 1 parenrightbigg n . Thus a n +1 a n = 3 parenleftbigg n n + 1 parenrightbigg n -→ 3 e > 1 as n → ∞ , so series (C) diverges. Consequently, of the given infinite series, only A and B converge. CalC7g23d 003 10.0 points Find the value of lim x → ∞ x 3 3 x . 1. limit = 0 correct 2. limit = 3 3. none of the other answers 4. limit = 1 3 5. limit = 6. limit = -∞ Explanation: Set f ( x ) = x 3 , g ( x ) = 3 x = e x ln3 . Then f, g are everywhere differentiable func- tions such that lim x → ∞ f ( x ) = lim x → ∞ g ( x ) = Thus L’Hospital’s Rule applies, in which case lim x → ∞ f ( x ) g ( x ) = lim x → ∞ f ( x ) g ( x ) . Now f ( x ) = 3 x 2 , g ( x ) = (ln 3) 3 x . But then lim x → ∞ f ( x ) g ( x ) = lim x → ∞ 3 x 2 (ln 3) 3 x , which, up to a constant, is the same limit we started with except that x 3 in the numerator has become x 2 . Consequently, if we apply L’Hospital’s rule sufficiently often, we finally end up with lim x → ∞ f ( x ) g ( x ) = lim x → ∞ constant 3 x . Thus lim x → ∞ x 3 3 x = 0 . CalC12b50s 004 10.0 points If the n th partial sum S n of an infinite series summationdisplay n =1 a n
Version 001 – tester – srinivasan – (54690) 3 is given by S n = 3 - n 2 n , find a n for n > 1. 1. a n = n - 2 2 n correct 2. a n = n - 2 2 n 1 3. a n = 3 parenleftbigg n - 2 2 n parenrightbigg 4. a n = 3 n - 2 2 n 5. a n = 3 parenleftbigg 3 n - 2 2 n parenrightbigg 6. a n = 3 parenleftbigg n - 2 2 n 1 parenrightbigg Explanation: By definition, the n th partial sum of summationdisplay n =1 a n is given by S n = a 1 + a 2 + · · · + a n . In particular, a n = braceleftbigg S n - S n 1 , n > 1, S n , n = 1. Thus a n = S n - S n 1 = n - 1 2 n 1 - n 2 n = 2 ( n - 1) 2 n - n 2 n when n > 1. Consequently, a n = n - 2 2 n for n > 1.

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