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Detemine wheiher the sequences are Increasing: decreasing or not monotonic If increasing enter as your answer If decreasing; enter as your answer: If not monotonic;...

Question

Detemine wheiher the sequences are Increasing: decreasing or not monotonic If increasing enter as your answer If decreasing; enter as your answer: If not monotonic; enter Nas your answer:1. On4 2. Un 3.Un An47 JR

Detemine wheiher the sequences are Increasing: decreasing or not monotonic If increasing enter as your answer If decreasing; enter as your answer: If not monotonic; enter Nas your answer: 1. On 4 2. Un 3.Un An47 JR



Answers

Determine whether the sequence is increasing, decreasing,
or not monotonic. Is the sequence bounded?
$$
a_{n}=(-2)^{n+1}
$$

So in this problem we have given Sequence described by the equation is seven is equal to two in minus three, divide by three and plus four. And what we wanted to find out here is to determine whether this sequence is actually increasing or decreasing or neither of the two. And also we also wanted to find out whether the the sequence this given sequence is founded. All right. So first thing I wanted to to do here is to transform this given ace event in such a way that I will multiply both the numerator and the denominator by one over you. One over N. I divided by in both the numerator and the denominator because in here is the term with the highest exponents. So a seven will now be simplified as two minus the over in Divided by 3-plus 4 over in and the reason why I did that is because we know that as approaches to infinity mm The value of one over an athlete approaches to zero. So that being sent as N approaches to infinity is and actually has value equivalent to approximately equal to to minus zero. All over 3-0 or simply equal to stick. Note that this these times. Here we already know that that will become zero as N approaches to infinitely. So that being said, a seven approaches the value of 2/3 as N approaches to infinity, which indicates that it is actually bounded right now let's do some quick check or a few values of a seven when n is equal to one, the value of SFM is 1/7 or approximately negative 10.14 when n is equal to its 1/10 or approximately equal 2.1. And then let's fight the blogging somewhere, values of N will end up having the following three over Um 13. So that's 13, which is approximately equal to .23. You have 5/16 here, which is approximately equal to .31 And then 7/19 which is approximately equal to point 37. And you can see here that it is showing an increasing pattern. So the answer to that question is that the answer to this problem is that this sequence is actually increasing And it is bounded because it will settle to a value of 2/3 as n approaches to infinity.

Let's consider the sequence given by N now. The first part is determine whether this thing is increasing, decreasing. Or perhaps it's not even Mon Atomic. Well, let's just write out a few terms here. A one that's just two minus one equals one. Now, how about a two? This one will be to plus a half. So five halfs. So from the first term, to the seconds her we saw increase, how about from the second terms of the third term? This time when and his three will have to minus the thirds, This is five thirds, and this is a decrease. So because of the term from a one to a two, this showed us that it was not decreasing because it increase. And from a two two a three, we see that it's not increasing. Therefore, the sequence A M is neither increasing nor decreasing, so it's not monitor tonic. Now let's go to the next part. We see that we start off at two, and then we're adding a term that goes to zero. So for this problem, it might be best to just noticed that the limit of an equals two so a N is bounded, and this is just coming from the fact that convergent sequences are bounded. This is an important that and we're using that to show and his bounded because it converges. And that's our final answer, not mon atomic, but it is bounding.

So that they're The sequence is written as and pictorial upon to power and right. So we can first of all we try to expand this sequence And and my next one and saw up to one upon. Sorry. Here it is hemisphere. So it will be an improvement. So this is canceled out. We can also write us and -1 factorial upon and So this and -1 upon and this value is separated and minus two of two. Mhm. one It is separated. This value is less than one and these old values are greater than All right, So for up to any quality too. Yeah, value decreases. And yes, here we can see here if any is greater than two will you increases Just we will write some values also to understand this. Okay, As zero is simply zero factorial upon zero Spirit. So it is not defined. So let's start from one as one will be one factory Lebron one Step 1 has two years two factorial upon to square, which is one x 4, one x 2 has three years 3 factorial upon. Too cute. Which is again six x 8. So we can see the value is decreasing here and again. The value increasing here after two Has by four. S whole factorial Upon two powerful, which is 24 by 16. So we can say that we lose increasing. So we can plot the values is starting from one, one by two Again six per year, 24 x 16 and so on. So this is the joining cold here. Right. Yeah. Mhm. No. Mhm. Yeah. You can see the graph is our sequences not monotonic.

So in this problem We have a sequence described by the equation. A seven is able to end all over in square platform and just a quick check, we can say that as an approaches infinity then this end right here. Any recent as N approaches to infinity? This guy here will approach to inventing as well and that is true as well for this term on the denominator it will also approach infinity. Only thing is that and squared plus one will become larger faster then in so n squared does one will approach to infinity faster than in which is denominator of this s of an equation. So what I'm trying to emphasize here is that if this denominator is getting more larger and larger as in increases then we could somewhat realize that is a N in general will become smaller as time progress or as in increases more and more and blinks just a quick check. So by simply analyzing the expression is f N. I am somewhat suggesting that this sequence should be decreasing but we can do some quick check here. So plugging in some few values of N. So when when N is equal to one will have one half here, 4.5 When any sequel to will have to over five here or .4 Then we'll have three over them here or .3 and then 4/17 here for roughly .23 and you see that yeah, it's true that the value of S F N is actually decreasing and the reason for that is because N squared plus one is increasing or approaching the infinity faster in a faster face than in in the numerator. So the denominator is increasing more fast, more quicker or faster than the numerator. Now the next thing we know now that it's decreasing, The next thing we wanted to find out is whether it is bounded or not. So what I wanted to do here is to somewhat right our is of an equation into another format by dividing both the numerator and the denominator by and square. And the reason why I'm dividing it with in squared is because um and squared here is the term with the highest exponent. So we'll have here one over in divided by one plus one over n squared. So a seven is now as N approaches infinity. Wait, before we do that, let's just recall that as N approaches infinity expressions like one over in and one over N squared are actually approaching to a value of zero. So that being said As in approaches in Finke seven will actually have a value of zero. zero for this one That will approach that will approach value zero as an approaches to infinity and the under the nominator side we have won a constant value and zero for this one. So simplifying further, we know that Okay, a seven will approach a value of zero as and increases more and more. So that being said, it is also bounded, it's decreasing and founded


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