could someone please explain and give answer Determine whether the series is abs

Question

could someone please explain and give answer

Determine whether the series is absolutely convergent, conditionally convergent or divergent. absolutely convergent divergent conditionally convergent

Explanation / Answer

1. This is not absolutely convergent. e^1/n > e^0 = 1, so e^1/n > 1. Thus, the sum of the absolute values is > the sum of 1/6n, which is divergent (you can use the integral test - the integral of 1/6n = 1/6 ln (n), and the lim n_> infinity 1/6 ln n = infinity) 2. We shall show that it is not divergent by showing that (3) it is conditionally convergent. 3. To show conditional convergence, we will show that it is alternating, the absolute value is monotone decreasing, and the limit of the sequence is 0. It is alternating because this is a multiplication of three terms, and only one of the terms is ever negative (e^(1/n) > 0 and 1/6n > 0). (-1)^(n-1) is alternating; it is positive for n odd and negative for n even. Thus, the product is positive for n odd and negative for n even, and is therefore alternating To show that the absolute value is monotone decreasing: As 1/n is monotone decreasing, e ^(1/n) is monotone decreasing, and 1/6n is monotone decreasing, so the product is monotone decreasing (terms are both positive and monotone decreasing implies the product is monotone decreasing). Finally, to show that the limit is 0. lim n -> infinity e^(1/n) = e^0 = 1 lim n -> infinity 1/6n = 0 Thus, lim n -> infinity Thus, lim n -> infinity e^(1/n)/6n = 0

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