In Exercises 121–124, determine whether the sequence is monoto...

Question

In Exercises 121–124, determine whether the sequence is monotonic and whether it is bounded.

aₙ = 2ⁿ 3ⁿ / n!

Accepted Answer

Hello there. Today we are going to solve the following practice prom 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. Let CN be equal to 5 power of n divided by n factorial. Determine if the sequence is monotonic and if it is bounded. OK. So it appears for this particular prompt, we are given a sequence and we're asked to ultimately determine if this sequence is monotonic and if it is bounded. So now that we know what we're ultimately trying to solve for, we need to recall a note based on our prior knowledge that in order to determine the properties of this specific sequence, we will first need to calculate the ratio of consecutive terms in order to see if they consistently increase or decrease and thereby as a result, we establish if the sequence will be monotonic. We will then need to examine the limits and behavior of the terms as n grows in order to determine if there are upper and lower bounds that the sequence will never cross. So we're trying to determine if there are upper and lower bounds that this specific sequence will never cross, which, which is never crossing. So with that said. Our first step that we need to take to solve this problem is we need to test for monoticity. So in order to determine if a sequence is monotomic, as we should recall, we will need to examine the ratio of consecutive terms which states that Rn, so. RN is equal to C subscript N + 1 divided by CN, and we will need to compare it to 1. So I'll make a note here and we're gonna compare it to one, so let's make that a note. So, we need to write that cn is equal to 5 n divided by n factorial, and the ratio of consecutive terms will be C subscript N + 1 divided by CN will be equal to 5 the power of n + 1 divided by parentheses of n + 1 multiplied by factorial. Multiplied by n factorial divided by drum roll 5 to the power of n. And this will be simply equal to 5 multiplied by 5 to the power of n divided by n + 1 in parentheses multiplied by n factor multiplied by n factorial divided by 5 to the power of n. And as we could see our 5 to the power of n. We'll cancel out with our 5 to the power of n, and our numerator and denominator will cancel, and our n factorial in the numerator and denominator will cancel, and that will leave us with 5 divided by n plus 1. Fantastic. So With that said, based on our prior knowledge, in order to be monotonic, the ratio must satisfy either, so let's make a note, so it needs to satisfy either. 1 CN plus 1, or rather C subscript N + 1 divided by CN is greater than or equal to 1 for all N. Or Or here's a big or C subscript N plus 1 divided by CN is less than or equal to 1 or all n, and once again n is some variable. So with that said. We need to note for our first case, we will consider N is less than 4, and if N is less than 4, that will mean that N is equal to 123. So we will then be able to write, so we need to say then. N + 1 will be less than 5, which will mean, so this will mean as a result that 5 divided by n + 1 will be greater than 1. Thus, we can go ahead and write that as a result, C subscript N + 1 will be greater than CN, and that will mean that as a result our sequence will be strictly increasing, so I put an arrow pointing up. For our second case, n is equal to 4, we need to note that if n is equal to 4, then that will mean that 5 divided by 4 plus 1 will be equal to 1. So thus, as a result, C subscript 5 will be equal to C subscript 4. For our third case where n is greater than 4, we need to note that if n is greater than or equal to 5. Then that will mean that N + 1. Is greater than 5. Which will mean, so we need to note that this will mean that 5 divided by n + 1 will be less than 1. Thus, as a result, we can write that C subscript N + 1 is less than CN, and that will mean as a result, our sequence will be strictly decreasing. So with that in mind, we can now make our final conclusions about the monoticity. And since the sequence is increasing, so it increases, or N is an element of curly brackets of 1.2, 3. We need to note that once again it's increasing for N is an element of curly brackets of 1,2,3. And we need to note on the other hand that it's decreasing for n is greater than or equal to 5. That will mean as a result, based on this information, it does not maintain a single change of direction, so it doesn't maintain a single direction of change for all n. So let's make a note of this for all n, which you'll have the symbol for, for all of. Which is like an upside down A, so for all N. So once again, this does not maintain a single direction of change for all N. So, therefore, based on this information, the series is not monotonic, so it's not. Monotonic, which I'll just put an M in quotes, so we know that it's not monotonic. So with that in mind, we now need to solve for the second part of our final answer, which we need to consider the proof of boundness. So based on our prior knowledge, we need to recall and note that a sequence is bounded if there exists a real number such as M, such that the absolute value of CN is less than or equal to M. Or All in. OK. So let's consider the lower bound first, and we need to note that since 5n is greater than 0 and n factorial is greater than 0, for all n is greater than or equal to 1. Then that will mean as a result that CN is greater than 0. So, therefore, 0 is a lower bound. So, 0 equals a lower bound, which I'll call LB for short. Considering our upper bound now. So, from our monoticity analysis, we can safely state, and you should know that the sequence increases until, so it increases until N equals 4, and it will stay constant at, so it's gonna be staying constant, which I'll put C in quotation marks. So it's staying constant at N is equal to 5, and it will decrease. For all N is greater than 5, and therefore this will, as a result, imply that the maximum value must be at, so we're gonna have our maximum value. B at N is equal to 4 or n is equal to 5. So let us state that M will be equal to the max of C4, C5, which will be equal to drumroll, 54 divided by 4 factorial, which will be equal to 625 divided by 24 when we write our final answer as a fraction. So that will mean for, so once again, for all N, so that will mean for all n. That will mean that CN will be less than or equal to 625 divided by 24. Thus, once again, you need to note that 625 divided by 24 will be an upper bound, which I'll call UB for short. So now we need to consider convergence, which will be the verification of boundness. So, we can also prove boundness by showing the sequence at, so once again, showing that the sequence converges. So, we can prove that there's boundness present by showing whether or not the sequence converges. So in order to do that, we will need to recall and use the ratio test for sequences which states that the limit as n approaches infinity of the absolute value of c subscript n + 1 divided by cn, which will be equal to the limit as n approaches infinity of 5 divided by n + 1, which will be equal to 0. And since the limit is L is less than 1, that means that the sequence CN converges to zero. So, let's make a note that CN converges to, so it's gonna be converges to 0. So it's converges to zero. So that means that every convergent sequence is bounded as a result. So, our final conclusions for the boundness will be that the sequence is bounded because 0 is less than cn and CN is less than or equal to 625 divided by 24, and this will be. For all N is an element of Z from the right. That's what the power symbol or the, the plus sign and the power sign means. So this plus sign and the power means. So that means that this will be true for all values of N, which is an element of Z from the right. So that means, so when we go back to the top to recap what we've learned, we can safely say that our final answer, based on all the information we've gathered together and our prior knowledge, we could safely say that our final answer is. Is, is that the sequence is not monotonic but is bounded. And that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 121–124, determine whether the sequence is monotonic and whether it is bounded.aₙ = 2ⁿ 3ⁿ / n!

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In Exercises 121–124, determine whether the sequence is monotonic and whether it is bounded.
aₙ = 2ⁿ 3ⁿ / n!