5

12, (ID points} Use the integral test to determine whether the series below converges or diverges. '5 In(n)...

Question

12, (ID points} Use the integral test to determine whether the series below converges or diverges. '5 In(n)

12, (ID points} Use the integral test to determine whether the series below converges or diverges. '5 In(n)



Answers

$3-8$ Use the Integral Test to determine whether the series is convergent or divergent.
$$\sum_{n=1}^{\infty} \frac{1}{n^{5}}$$

Right. We want to evaluate the convergence of the series. Using the integral test, the series in question is a sum from n equals onto infinity of 2/5 and minus one. Before we solve this problem, let's define the integral test to make sure we understand what we're implementing. So the integral test states that if the integral from K to infinity of F x dx converges so too must the sum from N equals K to infinity of AM. Similarly, if the integral from K to infinity fx dx diverges. So too must the series. So if we convert A. M to the form F X dx and take the integral from K to infinity, we can solve this problem. So that integral is from one to infinity to the x or five x minus one. This is the anti derivative to 55 x minus one. Evaluating from one to infinity gives two fists. Ellen infinity minus one minus Ln four. The first term goes to infinity. So the entire integral diverges to infinity. Since the integral diverges, that means are serious, diverse

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.

Okay, so problem 12. We are given this CVS some stars from to go into infinity off one over five and plus 10 times to apply the interval test. First, we want to see this Syriza's a function. So we put every backs equals one over five x plus 10 times x and we're going to first check these three criteria to use in general test. We're gonna first, um, it's, um check If ever relax is continuous on the domain, X equals two to the greater that to him of so on X equals or greater than two. Are these three gonna be satisfied? This is the first thing One check looking of dysfunction here. The only way for this fraction to the, um, invalid, I guess, is where we have zero in the denominator. But for all X equals two or graze in two, we will never have zero in the denominator. Hey, so yes, for all X equals two in grader Fo Becks exist. So it is continuous. Is it positive? Well, we know that the new mayor is positive for so if he ever get negative value, we gonna have some negative Ellen denominator. But for all X equals two or greater. We're gonna have 55 days and we always positive plus square root of X is also always positive. So 10 times positive valley is always positive. Therefore, denominator is gonna be always positive. Therefore, this whole fraction is also always positive. So the answer is yes. For all X equals or equal graven too. This fraction is always positive. Now is it decreasing? Um, weekend. Also, if we can force taken Davidov off off the function, see if he can find out if it's decreasing with the prison. But I think in this case, it's We don't have to do that because clearly five x is as exes gets larger and larger. Five X also gets larger and larger as excuse. Larger and larger square root of edges also get larger so 10 times something that gets bigger and bigger off course, biggest larger and larger. So we know that as X as ex increase the denominator five x plus 10 times x 10 times The square root of X also gets larger and larger. Therefore, one over, um, this guy right 50 x five x plus 10 times square root of X gets smaller as X increases. So, yes, the function is decreasing. Okay, so we verify that on this domain, all these three criteria are satisfied. So now we know that we can pile into protest. So let's take a interval. Starting from two to infinity. Off this function F of X, which is one over five x plus 10 times screening. Becks, DX. Okay. This is to violate the limit off this arbitrary bound be going to infinity. They were gonna said the bound from two to this object Rebound off one over five X plus 10 times. Good. Rex DX. Okay, We want to take limit of this interval. All right? How can we do that? Oh, all right. So let's put I'm gonna be right this one more time. Let's pack two out of five. Limit of big one to infinity. To be one over five. I fucked out five. So I'm gonna have X plus, uh, two squared of X. The ex Uh oh. You know what? This factor out five. Scared of X. That's factor out. Five squared of eggs. Common factor. So I'll have squared of exterior plus two. Okay, So now used a U substitution for squared of X plus two. Okay, then I have d'you over DX. I'll have s o This is extra the negative x to the 1/2. So, um, two squared of eggs. I want a numerator. Then what I want to get is DX over five times square of X. So how can I get that? Well, first, I'm gonna manipulate this too. Okay. T U equals DX over too scared of X. So let's multiply bulls size by, uh, two of five, do you? Then I'll have five squared of eggs and a denominator and then have DX Okay, this is what I wanted DX over five times quarterbacks. All right, so now let's rewrite everything in terms of you instead of eggs. So limit be going to infinity of all right When X equals two, I have squared of two plus two as you and the upper bond would be scared of B plus two. Come then, this whole expression I have, uh so one over. We put that issue times DX over five. Spirit of X is do over five. Do you? All right, So what I have is limit be going to infinity of this list is take this constant outside of the interval, but two over five times square root of two plus two skirt B plus two won over you, Do you? We can go ahead on complete this. So one over you is a ln all of you. And we go on developing this from on these grounds. Okay, Going to exchange. Let's just let me let me rewrite what I just wrote in the last lines. I was limit beagle into infinity of Ah, it was 25 It's too far. Ellen of you. L of you and the bounds were screwed. A group plus too scared of B plus two. So two over five, uh, limit hopes. All right. So since his constant I took taking outside of the mets, so then it I want to be a wanton community of now I'm gonna plug in these bounds. So ln off square root of B plus two minus ln off squared of two plus two. Okay, but this is just a constant and be gets larger and larger square root of B gets larger and larger. So, of course scared or B plus to go into infinity as be going to infinity. So elan off something going to infinity because Alan is increasing function because logic, logic. So after all, this whole thing is going to get larger and larger going to infinity. Therefore, says this even an integral I was too going to infinite off this original first. If OPEC's DX is the version, um therefore the Siri's is diversions, and that's the end of the argument.

Using the integral test of the Siri's. We'll take the limit a Z. He goes to infinity of the integral off one from one to be. And then we're going to use the power rule on the outside here. So this would give us a new power of N plus four to the one half and then divided by one half of multiplied by two. Then we don't need to worry about you substitution, since the derivative of N Plus four is just one. So I would just divide that whole expression by one. And then we're going to evaluate this from one to be. And of course, we're still letting the approach infinity. And so therefore, this would leave us with infinity in the mix here. And of course, that that's going to diverge off to infinity. And it doesn't matter what the rest of it is. So therefore, the original integral or excuse me. The original series is divergent by the integral tests since the integral diverged off to infinity


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