11–86. Applying convergence tests Determine whether the follow...

Question

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.

∑ (from k = 1 to ∞)k! / (kᵏ + 3)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says, does the series, the sum of M equals 1 to infinity of M factoral divided by m rates to the quantity of M + 2 and quantity converge. And we're given two possible choices as our answers. For choice A, we have yes, the series converges, and for choice B, we have no, the series diverges. So in order to determine whether our series here converges or diverges, we're going to be using the ratio test. So recall for the ratio test that we want to calculate L, which is going to be equal to the limit as M approaches infinity of the absolute value of the quantity of a subm plus 1 divided by a sub M. In quantity. Where a sub M in our case, is going to be equal to M factoral, divided by M raised to the quantity of M + 2 and quantity. So using that definition for A of M, that means that L is going to be equal to the limit, as M approaches infinity of the absolute value. Of the quantity of the quantity of M + 1 in quantity factoral divided by the quantity of M + 1 in quantity raises to the quantity of M + 3 in quantity, and that's going to be divided by The quantity of M factoral divided by M rates to the quantity of M + 2 in quantity. So, next we want to simplify this expression. So, this is going to be equal to the limit. As M approaches infinity of the absolute value of the quantity of the quantity of m plus 1 in quantity multiplied by M raises to the quantity of M + 2 in quantity divided by the quantity of M + 1 in quantity rates to the quantity of M + 3 in quantity. And we could continue to simplify this further, and so this is going to be equal to. The limit as M approaches infinity. Of M rates to the quantity of M + 2 in quantity divided by The quantity of M + 1 in quantity rates to the quantity of M + 2 in quantity. And so here we drop the absolute value signs since M here is a positive value and so therefore both M and M + 1 are positive values. So, next we can just rewrite this as being equal to the limit, as M approaches infinity. Of the quantity of M divided by the quantity of M + 1 in quantity, that whole fraction is raised to the quantity of M + 2 in quantity. So now, instead of trying to take this limit, we're actually going to look at the natural log of L. And so the natural log of L is going to be equal to the limit as M approaches infinity. Of the natural log of the quantity of M divided by the quantity of M + 1 and quantity in that whole fraction is raised to the quantity of M + 2 in quantity. And we use properties of logarithms to rewrite this as being equal to the limit. As M approaches infinity of the quantity of m plus 2 in quantity multiplied by the natural log of the quantity, and here inside the argument of the natural log function will divide both the numerator and denominator by M. So we'll have 1 divided by the quantity of 1 plus 1 divided by m and quantity. So next we want to do is to break this up into two different limits by distributing the natural log of L into the quantity of M + 2. In doing so, this is going to be equal to the limit as M approaches infinity of M multiplied by the natural log of the quantity of 1 divided by the quantity of 1 plus 1 divided by M in quantity, then we'll have plus the limit as M approaches infinity. Of 2 multiplied by the natural log of the quantity of 1 divided by the quantity of 1 plus 1 divided by M and quantity. Now we want to do is to look at these two limits independently. So for our first limit, We're gonna have the limit. As M approaches infinity. Of, and here we're going to rewrite. Um, our expression here. So we're going to use the properties for logarithms to rewrite the natural log of the quantity of 1 divided by quantity of 1 plus 1 divided by m in quantity as minus the natural log of the quantity of 1 + 1 divided by M in quantity, and this is going to be divided by the quantity of 1 divided by M. And the reason why we wrote it in this form is that recall that the natural log of the quantity of 1 + X in quantity divided by X, that if we take the limit as X approaches 0 of this quantity, that this is equal to 1. So if we set X equal to 1 divided by M, then as M approaches infinity, X is going to approach 0. So that means that we can use this expression to determine our limit. So that means that the limit as M approaches infinity of minus the natural log of the quantity of 1 plus 1 divided by M in quantity, divided by the quantity of 1 divided by m is going to just be equal to -1. So next we'll look at the 2nd limit, which is going to be the limit as M approaches infinity. Of 2 multiplied by the natural log of the quantity of 1 divided by the quantity of 1 plus 1 divided by m in quantity, and as M approaches infinity, 1 divided by M approaches 0, so that means that this is going to be equal to 2 multiplied by the natural log of the quantity of 1 divided by the quantity of 1 + 0 in quantity. And this will have the natural log of 1 multiplied by 2. We call the natural log of 1 is 0, so that means that this whole quantity is going to be equal to 0. So that means that the natural log of L is just going to be equal to -1 + 0, which is going to be equal to -1. So if we exponentiate both sides of this expression, we get that L is going to be equal toe raised to the quantity of minus 1 and quantity, which is equal to 1 divided by E. Now recall that if L is less than 1, then our series converges. And if we look at L, which is equal to 1 divided by E, that is indeed less than 1. That means that our series converges. And since our series converges, that means that the answer to this problem is going to be yes. So, if we compare our answer here to our answer choices, You'll see that the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞)k! / (kᵏ + 3)

Key Concepts

Convergence of Infinite Series

Ratio Test

Factorials and Exponential Growth Comparison

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