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Is the statement "the integral test is applicable to the seriesTRUEsinYes...

Question

Is the statement "the integral test is applicable to the seriesTRUEsinYes

Is the statement "the integral test is applicable to the series TRUE sin Yes



Answers

Can the Integral Test be used to determine whether a series diverges?

What he called a consent under been to go fast. It's saying that if I die sponsored video and the next time Miss Mama on the culture that gave us time for all guy And from here we have to be heavy. I'm the submission on. Okay? And then MP has Cem. Is that the girl off from one to infinity? I asked the banks and they're far suppose done this to go here and then if the IndyCar one to infinity and I angst the ex coach in symmetry isn't obliged. Under submission on that, I can watch Infinity. Absolutely coach insanity. And it may never be divergent. Now, far, this estimate in the question here would be true.

The integral test with this here. So All right, checking to see why the integral test doesn't work with this actually. So we're going to go ahead and take the dread over this. Take a look at that. We have to sign up and no go ahead and apply the kool Cheryl after that. So it's going to be a G F prime just and co signing event is F cheap primes that sign him in times one overhead squared. Okay. So if we take a look at that here, so can already see that end. Cause I never end minus sign event, right? It's not always negative. Even when we multiply that by the sign of in there, that component there too. It's the same thing. So that's being multiplied here. That so therefore cannot use the integral test.

So we're trying to see here why we can't use the integral test on this series here. And so something that we can note here is that we've got sign event and the very nature of sign of hand here because I mean it goes up and down everything. We've got the plus two aspect here. So that means that side event is now going to go between one hand three. So that's okay here. But the thing we need to note here is that the derivative like this. So let's take the trade of this year. So we've got G. F prime. So that's at times co sign event minus Yes, two plus sign in this one all overhead squared. So then that leaves us with so and CO san I have this to minus side event over and squared. So we see that we've got co sign and sign on top here and squared on the bottom is always positive. But the thing is that was so with sign and co sign on top here. Everything this is not always not always decreasing. So therefore we cannot use the integral test

Given that this Siri's converges and and positive. Now we'd like to see if the serious convergence since the limit of an well, actually, let me take a step back here. So since converges, this limit has to be zero. This's by the test for diversions. Come on. So there exists and end such that if little and his bankers and begin than zero Weston am less than one. So now, with use the fact that okay, sign X is always less than or equal to X. This is if X is positive. So this implies Signe and less than or equal to an and therefore if little and is bigger than capital and then sign of an It's just some number between zero and one, as long as the input, the angles between zero and one so we can write this sum. It doesn't matter what the starting point is like his aides for more and equals one. No, we can write this now on the second. Siri's over here. We know that this Siri's will converge because this is less than or equal to just an and then we have just a finite sum here. So when we go on to the next page. So this is the whole right hand side is less than or equals who this converges. This is just a finite sum. So it emerges and therefore the Siri's converges by the comparison test. So let me make a remark here, although it is possible that now all of these guys who are bigger than zero what we showed was that by writing it in this form that eventually it is bigger than zero. And that's what really matters here. This is why we split it into two sums and then on the second some here and is bigger than capital in. And this is from what I showed in the previous page. I know that sign was positive for these values of them, and so, really, I used the comparison test on this. Siri's only So this is a finite Siri's and that this over here is just a finite sum. And if I add these two together, the result is that it convergence by the computer success, and that's the final answer


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