55–70. More sequences
Find the limit of the following se...

Question

55–70. More sequences

Find the limit of the following sequences or determine that the sequence diverges.

aₙ = (−1)ⁿ ⁿ√n

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the limit of the sequence Bn, which is equal to the quantity of minus 1 in quantity rates to the power n multiplied by the nth root of the quantity of 3 N in quantity as n approaches infinity. And we're given 4 possible choices as our answers. For choice A we have 1. For choice B, we have 0. for choice C, we have minus 1. And for choice D, we have the sequence does not converge. Now we're asked to find the limits of our sequence as N approaches infinity, and we're given the sequence B N. Which is a product of two quantities. So we're gonna begin by focusing on the 2nd term in our product, and we're gonna be looking at the limit. As in purchase infinity. Of the nth root of the quantity of 3 N. In quantity, and we can rewrite this as being equal to the limit. As in purchases infinity. Of the quantity of 3 N in quantity rates to the quantity of 1 divided by N in quantity. Now, if we try to take the limit in this form, we'd get infinity rates of the power of zero, which is an indeterminate form. So we're actually going to do is look at taking the limit as end approaches affinity of the natural log of this quantity. So if we look at the limit, As in per infinity. Of the natural log of the quantity of 3 N in quantity raised to the quantity of 1 divided by N. And this is going to be equal to the limit. As N approaches infinity of 1 divided by N multiplied by the natural log of the quantity of 3 N in quantity. Which is just equal to by rewriting. Our expression here, the limit as in approaches infinity. Of the natural log of the quantity of 3 N in quantity divided by N. Now, if we try to take the limit as N approaches infinity, we get infinity divided by infinity, which is an indeterminate form, but it is a form that which in which we can use Opital's rule on. So that means that this is going to be equal to. The limit As N approaches infinity, and we call for Lopita's rule, we need to re-evaluate our limit after we've taken the derivative of both the numerator and denominator with respect to N. And so when we take the derivative of our numerator, we will get. 1 divided by the quantity of 3 N in quantity multiplied by 3, then that quantity is all divided by the derivative of a denominator, which when we take the derivative denominator with respect to N, we get 1. That means that this is going to be equal to the limit as N purchase infinity of 1 divided by N, which when we evaluate this limit, as it approaches infinity, quantity of 1 divided by N approaches 0, so that means that this is going to be equal to 0. So that means that the limit As in approaches infinity. Of the in fruit of the quantity of 3 N. In quantity is going to be equal to, so we just take our previous expression, which is going to be the limit as in approach infinity of the natural log of the quantity of 3 N and quantity raised to the quantity of 1 divided by n in quantity and exponentiate both sides of that expression. So we'll have this limit being equal to e 0. Which is just equal to one. So, now we need to look at the first terminal product of our sequence B sub N. And if you look as in approach is infinity, We see that the quantity of -1 in quantity rates to the power N. does not approach a single value, but instead it alternates between two values. So it alternates. Between the values of -1 and 1. So the quantity of 1 in quantity of parian alternates between -1. And one. So it does not approach a single value. So that means that as in purchases infinity. That means that our sequence, which is B N. We will be alternating. Between two values as well. So this event alternates. Between the values of -1 and 1 as well, since the limit as in the approaches infinity of the second term in our product approaches 1. And so since BN does not converge to a single value, that means that this sequence does not. Converge. So that is the answer that we're looking for for this problem. And if we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice D. So I hope that this explanation has been useful, and I'll see you in the next video.

55–70. More sequences
Find the limit of the following sequences or determine that the sequence diverges.

aₙ = (−1)ⁿ ⁿ√n

Key Concepts

Limits of Sequences

Nth Root and Its Limit

Alternating Sequences

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