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(20 Points)Use Integral Test to decide if the series[arctan (2n)]2 An2 + 1converges Or diverges_Find the SUM of the series(2n 1)(2n +3)...

Question

(20 Points)Use Integral Test to decide if the series[arctan (2n)]2 An2 + 1converges Or diverges_Find the SUM of the series(2n 1)(2n +3)

(20 Points) Use Integral Test to decide if the series [arctan (2n)]2 An2 + 1 converges Or diverges_ Find the SUM of the series (2n 1)(2n +3)



Answers

$21-34$ Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum_{n=1}^{\infty} \arctan n$$

Can discussion a series is given that is submission and it calls to 12 in finite and divided by any square plus one. Okay. And now we have to decide whether the series converges or diverges using that integral test. Okay? So first of all I am in the cities is and divided by any square plus one. Okay. And two integral test. First of all, we will take the function F X equals two. If X equals two X divided by X square plus one. Okay? But before applying the integral test we have to check whether the series is decreasing and positive or not. Okay? So for a series decreasing the submission equals and equals to one point finite. Am should be equals to zero. Okay? And the series is positive we can say and 4°.. We will test this. So it will be integration and equals to 12 in finite and A N. S. And divided by any square plus one. Okay. And when we saw it will be integration and equals to one in finite. And we will take and common factor so it will be one divided by and bless one by n. Okay. And when we apply the limit here this will be zero and it will be in finite and one by in finite that will be zero. Okay. So we can say our series is decreasing and it is positive. So it is safe to apply the integral test here. Okay. And now our function is X divided by X squared plus one. So for integral test integration of F X D X. That is X divided by X squared plus one dx from one to in finite. Okay. We have to check this F And now in this integral test three and 9.3. Okay. If this integral converges so our series also converges and if this integral diverges over series also diverges. Okay, so we will solve this integration So it will be okay integration of here 112 in finite. So the integration of X divided by X square plus one. It will be Ln one by two. L n x square plus one. Okay. And the limit is here +12 in finite. Okay. And now when we apply this it will be mm one by two. Okay. First a parliament that will be in finite and minus lower element. It will be won by two. Ln two. Okay. Or we can say this will be infinite minus some value. That will be in my night. So here this integral diverges. Okay, so we can say the integral diverges over series will also diverges. Okay, so I'm writing down the final answer here diverges and this will be the final answer of discussion. Thank you.

Okay, discussion series is given that is submission and it calls to you in finite one divided by N L N N. Holy Square. Okay. And we have to find out we have to decide whether the series converges or diverges using that integral test. Okay, first of all I am is here one divided by N L N N. Holy Square. Okay. And we have to apply the integral test. So integral test only can be applied if the function is decreasing and positive. Okay? So it is positive, we can say but the decreasing we will test integration and it was 222 in finite and it should be called to zero. Okay? So here submission and it calls to to to and finite one divided by N. L. N. N. Holy Square. Okay. And when we apply intends to infinite hair, so it will be won by infinite and it will be zero. So that we can say this function is decreasing and also positive. So we can apply the integral test here. Okay, so A N He is one divided by an L N. In holy Squire. So over function effects will be one divided by X. Let X Holy square. Okay. And to use the integral test the integration of f x D X. Okay. And the limited to two in finite it will be integration to to in finite one divided by X. Length X. Holy square and dx Ok. And now if this function if this integral converses so over series will also converges and if this integral diverges then our series also diverges. Okay, this is the integral test. And now to integrate one by X L N X plus in L N X. Holy Square the integral will be minus one divided by Ln X. Okay, you can solve this integral separately by substituting U equals to L N X. Okay then the derivative one by X. Dx is also there and it will form one by us square And the integration will be minus one by you. That is minus one by L N X. Okay, I have right down the integral directly. Or you can solid separately. And now the integration it is to to infinite. Okay, that can be written as it will be too in finite. Okay, and now When we apply this it will be uh -1 divided by in finite. Okay. And it will be negative and -1 divided by island. To okay, so it will be zero and negative negative positive one divided by island too. Or we can say one divided by Ellen too. And it is a definite at bound and bounded answer. So we can say this integral converges. OK. And if over this integral converges then our series also converges. This is the Durham 9.3. Thank you.

We want to use the integral test to determine whether this series converges or diverges, and we want to check that the conditions for the interval tests are satisfied. So things I'm gonna do is I'm gonna first factor out of five from this because if we factor out a constant, this infinite series still is gonna converge or diverge regardless. So we're gonna have 1/5 integral from And is he going to to infinity of so one over and plus to root? And working with this function will just be a little bit easier, since we don't have that extra five floating around in the denominator. But now let's go ahead and check to see that we can actually apply Inderal test. So I'm gonna let f b 2 1/2 plus two route head and well, this should be strictly greater than zero for all values where this is defined. Since square rooting a positive number gives us a positive number multiplied by Tuesday. Positive ad positive, Positive. So this is going to be strictly greater than zero. And then this is also continuous from in his gratitude to infinity, since the square root is going to be defined. And then if we're adding to positive numbers, we never get zero. So we don't have any issues with that. Now, let's go ahead and check to see if it's decreasing so we can take the derivative of this. So I was going to use questionable to take the derivative because I don't want to use power rule, so it's gonna be low de high, the derivative. That's just gonna be zero and then minus high, de Lo and the derivative of that should be one plus one over the square root of in, and that's using power. And then over what we have in the denominator squared, right? So that term just goes away, and so we have one plus one over route in, which is gonna be positive in plus two over route and square. That's all. It is going to be positive, and that negative there makes it so that this function is always less than zero. So seems like everything checks out so we could go ahead and integrate that. So I'm just gonna integrate this. We're gonna pretend like that 1/5 isn't there just to kind of simplify the calculations so we're going to have the integral oh to to infinity of one over so would be in plus two routes. And I'm gonna do something that might seem a little bit un intuitive. And I'm gonna factor out that route in, so I'm gonna have roots in Fruit n Plus two dia. Now, the reason why I've been want to do this is because now we have a substitution that we can apply, and there might be a better way to go about this. But this is the first way that I kind of thought about integrating this. So we're gonna let you eagle to root in Plus two, taking the derivative this gives us, Do you? Is it good to one over to root in Deon? So we just need to multiply and divide dysfunction by two. So tattoos and the denominator here we multiplied by two, cause that's just multiplying by one. And now we have our d. So let's go ahead and plug. All that didn't know. So our new bounds of integration should be well, if we plug in to as going to give us Route two plus two and then if we take the limit as n goes to infinity or this that should so go to infinity and then we're going toe have just one over you, do you now integrating that will give us the absolute value of the natural log or the dacha log of the absolute value of you. And we evaluate from route two plus two to infinity. Well, we plug in affinity or if we take the limit as you goes to infinity, then that diverges to infinity. And then we subtract off the natural log a two plus two. But I mean, that doesn't change anything. So this interval still diverges. So that implies are Siri's. And is he going to to infinity of 1/5 plus 10 Route five in plus 10 room in also Die Burgess.

So we have here a positive continues decreasing function, which means we can use the integral test. So we'LL take the integral from one to infinity to over five X minus one t x And of course, we need to change that into a limit and we're gonna use you substitution to do this integral. So we'll use the substitution with u equals five x minus one, which would make do you five d x. So this is really equal to the limit The studio's turn affinity any girl from one to infinity Or really, because we're changing the bounds, we should say so. Kind of a to b of too over you times to you over five The do over five comes from dividing the five from r D X over to the d you side. Now we can integrate giving us two fifths Allen of the absolute value of you. Ah, for the bounds of our into girls and replacing our ex right peace back in for you. We have this for integral. Now when we actually plug our bounds of integration, would that be should be a t We find that we end up with I two fifths, Alan five T minus one minus two fifths. Fell into four. No. And as t goes to infinity, this whole thing goes to infinity, so we know the Siri's is diversion.


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