Suppose that aₙ 0 and limₙ→∞ n²aₙ = 0. Prove that ∑aₙ con...

Question

Suppose that aₙ > 0 and limₙ→∞ n²aₙ = 0. Prove that ∑aₙ converges.

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Suppose ann is greater than 0 and the limit as n approaches infinity of n to the power of 3 divided by 2 multiplied by a n is equal to 0. Determine whether the series, the sum evaluated from n equals 1 to infinity of a n converges or diverges. OK. So it appears for this particular prong. We are given information. We are told to suppose a n is greater than 0, and we're told that the limit as n approaches infinity of n to the power of 3 divided by 2 multiplied by a n is equal to 0, and based on this information. We are asked to determine with the series that is provided to us, we need to determine if this series will either A converge or B diverges based on all the information that is provided to us. So now that we know that we're ultimately trying to determine if the series given to us by the prom itself is converging or diverging. Our first step that we need to take is we need to state what our given limit is. So once again we are told that the limit as n approaches infinity of n to the power of 3 divided by 2 multiplied by a n, specifically a subscript n is equal to 0. And we will need to recall and use the definition of a limit to choose capital N, which is some integer value, so that all lowercase n is greater than or equal to N, and that will mean we will have n to the power of 3 divided by 2 multiplied by and will be less than 1. Thus, AN will be less than 1 divided by N to the power of 3 divided by 2. And this will be for all lowercase n is greater than or equal to capital N. Therefore, we will have 0 is less than an N. And AN is less than 1 divided by n to the power of 3 divided by 2. So at this point, we need to recall and apply the direct comparison test, and as we should recall, the direct comparison test states if 0 is less than or equal to AN and less than equal to bn. For all n, so once again, if 0 is less than or equal to Ann and Ann is less than or equal to bn, for all n and the sum of bn converges, then the sum of an converges. Conversely, if 0 is less than or equal to n and bn. So once again, conversely, if 0 is less than or equal to Ann and Ann is less than or equal to bn for all lowercase n, and the sum of a n diverges, then as a result, the sum of bn will also diverge. So with that in mind, we need to recall and recognize that the series, the sum evaluated from n equals 1 to infinity of 1 divided by n to the power of 3 divided by 2. We need to recognize that this is a convergent P series, and this is due to the fact that sense P is equal to 3 divided by 2, which is greater than 1. Therefore, as a result, the sum evaluated from n equals 1 to infinity of a n will also converge. So that's our final answer, is it converges. Hooray, we did it. We solved for this particular problem. So going back to look at our multiple choice answers, we could safely say that our final answer once again is multiple choice answer letter A converges, and that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Suppose that aₙ > 0 and limₙ→∞ n²aₙ = 0. Prove that ∑aₙ converges.

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