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N4 Sn' + 2 Drtcrmino M thc soqucnce converges, If so find the lumnit 7n8 + 8n8 + 5 "3...

Question

N4 Sn' + 2 Drtcrmino M thc soqucnce converges, If so find the lumnit 7n8 + 8n8 + 5 "3

n4 Sn' + 2 Drtcrmino M thc soqucnce converges, If so find the lumnit 7n8 + 8n8 + 5 "3



Answers

Determine whether the sequence converges or diverges. If it converges, find the limit.
$ a_n = \frac {3 + 5n^2}{1 + n} $

To figure out whether or not this converges. We just need to figure out whether or not this limit exists. Okay, so we could do the trick where we divide the top and the bottom by the largest power of end that we see in the denominator. It's in this case we'd be dividing the top in the bottom. Bye. In squared, as in Goes to infinity three over and squared, goes to zero and one over and goes to zero as well. So we just have five over one, which is five. So it does converge and it converges two five.

So in this problem we have a sequence described by the equation is event Is equal to 3 to the end. Post to all over five. Race to end. So let's do a quick check before we answer this problem. So basically what we wanted to find out in this problem is whether this sequence will convert to a certain value or not And doing quick chocolate substitute the first few value of end. So let's say when any she got 1/8 of end should be equal to three cues all over Hold on, let me try to make that a little bit smaller. So that should be equal to treat cube all over all right, Or 27/5. Ultimately private plane porter when she got two will end up having three reach to four over five squared and that should be equal to um 81/25 Or approximately 3.24 When N is equal to three when this becomes 32 50 and over five Cube were approximately equal to 243. All over 125 which is equivalent to 1.944. When N is equal to four, we'll have three races the 6, 1 over 5-4 Equal to 729. All over 625 Were approximately equal to 1.1664. We're sure that the value is getting smaller and smaller, but what is the limit of this value As it gets smaller, smaller. So let's try to manipulate a little bit the equation of s a pen to identify the limit and let's see whether it is converting to a certain value or not. At this very moment we could somewhat say it will converge but what is the limit of that? What in what well will it converge as N approaches to infinity. So what I wanted to do here is that instead of having this, take note that this street based on the end of story can be written as street to the end frames, peace groups divided by frightening from what I will. What I wanted to do here is to multiply viable denominator and denominator by one over. We will be in and multiply the nominator escalate with 1/3 to the end. I'm doing that. I would end up having and the nominator I will end up having please forward. So just to start here, we're just canceled out Leaving an value equal to one and the denominator will become five To the end. All over 3 to the end. So let's revive it this way. nine over five thirds race to the end Signal 5/3 is a value more than one. All right, So as n approaches to infinity 5/3 will just grow and grow more and more. five Judge speaks to be in will just grow more and more and will approach infinity Right? Any number that is created in one. As you try to base it to a certain exponents and that exponents becomes larger and larger, the equivalent value of that fraction or that part of the or that base will just become larger and larger. So I could simply say that as N approaches infinity. 5 3rd space to the end approaches to infinity as well. That being sent ace of em as N approaches to the infinity will just be equal. Take note. This is the equivalent expression we created. That is equivalent to this one, So is seven is simply equal to nine over infinity Because we know that 5/3 face to be in is only approaching a value of infinity as N approaches infinity. In any number divided by infinity except zero and infinity Is equal to zero. That being said, let me write these like let me write the answer this way. So this sequence converges 20. It converges to a value of zero.

Even sequences and plus sign of em. And on the denominator side we have costs for and rape. So related funds and hair will be X plus sine of X upon X plus Korsakov four X. So for limit acts approaches infinity This term will approach to X plus sign off X upon X plus course of four X. Right. So you're using hell hospitals rule. We will have limit X approaches infinity differentiation of that nominated part Which is one plus course of X. And on the Sorry this is name later. And on the denominator we will have minus sign off X for X into four. This will be limit X approaches infinity one plus core sex 11 minus four cynic forex. This is one plus or soft x. So this time will near at infinity we cannot say the value but we can say definitely say since cause X and cynics belongs to 0 to 1. Sorry, minus 1 to 1, ranging minus 1 to 1. So we can say this time is finite. Mhm. Fine item. Because core sex belongs -1-1 this value. So we can say the graph converges the graph or sequence, and we're just yes.

In this problem, we wish to determine the convergence of the given sequence A. N equals five, divided by N plus two. This question is challenging your understanding uh sequences and how to determine whether or not a given sequence converges in order to solve the way we're going to approach is by using the limit. So we applied a limit to the value end of the sequence limited and approaches infinity A. N. We see that the limit as N approaches infinity, A. N equals L. The sequence will converge to value L. And if the limit does not exist, the sequence diverges. So if we need to limit and approaching infinity and fucking infinity to R. N, we can evaluate what this sequence converges do and determine whether or not it is convergent. So limit as N approaches infinity of our A. M. Is fiber and plus two equals fiber. Infinity plus two equals five over infinity. Remember the dividing a non infinite number by infinity means it goes to zero because it becomes infinitely small. That's a solution of zero. Thus we confirm That are given sequence converges to value zero.


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