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Find the values of p for which the series is convergent: Enter your answer as an inequality.)n = 2 n(In(n)) P...

Question

Find the values of p for which the series is convergent: Enter your answer as an inequality.)n = 2 n(In(n)) P

Find the values of p for which the series is convergent: Enter your answer as an inequality.) n = 2 n(In(n)) P



Answers

Find the values of $ p $ for which the series is convergent.
$ \displaystyle \sum_{n = 2}^{\infty} \frac {{1}}{{n(\ln n)^P}} $

Even industries on the form one off in terms on an under an hour p and was from to infinity here I will apply the Interco test here on We need to compute the integral out of form one on the X Times and another ex about API does from to infinity, the ANC's and here in this case, and will use the new ICO to a line of X. Therefore again d'you equal to one of thanks, the X and I would chance the limit as well. So it's you again the, uh you ex echo to chew and then we get a new culture Ellen up to X pickle to infinity and then began the new equity infinity. Therefore, I should be able to find this one. Headaches certainly will be. Did d you hear on that we should rewrite the interviewer is the lead up to goes to infinity and I'm gonna d you over you about PPI so we can get on as the u Belmont's p. D. You from an end of a Jew to infinity. Now, in this case, we will consider two cases here for the be equal to one and for the B not equal to one. And then we do in this in discs it let me write down in here and now for the be equal to one. And for that be not equal to one for the big could you won again the co equal to the land off blood you Andy violated from and then up to to infinity. And therefore we should ganda when you but infinite inside again the infinity manners and land off on an actual So this Numico choose infinity. Therefore, bad Interco test hand a series here will be divergent And now for the second case, when b is ah, it's not echo juice Era is not equal to one with spinach you that we will have the do you integral becomes a U boat after minus b I plus one dividing by a man It's be best one and then evaluated from on a two to infinity and we see that this one we're equal. Do the ah It would have the b hair and will be credit and one and then every go juice. Ah, zero and then plus the island A Jew about I'm the A minus bi bless one be running by the minus B bus Kwan If only If there be here greater than one and it will go to infinity, even be here smaller than one. As a result, we get that disagrees Here it will be convergent even only if be here must be credit and one

Let's use the Inderal test and we'LL see why this works out nicely in a second. So just replace and with the ex and first thing is a cz usual. You could just write this as a improper, integral and then if it helps, you can also do a use up here. Now the steppe is not mandatory, but it'LL require less work. So I'm just going to convert this entire integral into the new variable you So notice here, this was X equals one. So plugging that in for X over here, you see that you have one equals two. That's our new lower limit. And similarly, if you plug in the upper bomb, be in for X, you get our upper limit b squared plus one and there's the rest of our internal rule. So now we should consider some cases here, case one p equals negative one. And in that case, the integral becomes ln absolute value over too. But when you plug in the B squared plus one into the log and take the limit, this will go to infinity. So diverges so there. It means that we do not want to consider p equals minus one. Let's go to the next page here. So now we consider P not equal to minus one. And when we do that, we can go ahead and use the power rule to evaluate the integral. This is why we're considering two cases because the anti derivative is different, depending whether or not pee is equal to negative one. So now we're plugging in, and so at this point, we need to consider cause notice that one plus b squared goes to infinity. So we should consider whether or not this exponents is positive or negative. If P is less than negative one, then P plus one is less than zero. And this will ensure that now we have a negative exponents. This will ensure that the infinity goes in the denominator and that will end up being a zero. So if P is less the negative one, the integral just becomes this term here, which is a real number. You're not dividing by zero so converges in this case, and then consider the other case he's bigger than negative one. This implies P plus whoops. Sorry about that. This implies P plus one is bigger than zero. But that only mean that the infinity has a positive exponents and that will ensure that the entire in a girl is infinity. So, out of all the cases, the only one that we had conversions was a P was strictly less than negative one.

So here we'd liketo find what P values make this Siri's conversion. So let's apply the integral test so we can define after Becks as Ellen X overexert the P. And here you see Ann is at least one. So we'LL take Ex to be at least one now. To apply the integral tests, we have to make sure that the hypotheses air satisfied. So the first one ethics is positive if X is bigger than one, and we can see this because natural log in exit appealable positive. And if you divide two positives, it's still positive. Further, we need that affects is monitoring decreasing? And perhaps the easiest way to show this message is compute derivative of F and show that it's negative. So here you would just use the potion rule and then show that it's negative. If X is larger than one. And if you have to, you apply abound on P. That's fine, too, but you shouldn't have to hear. Okay, now, using the integral test, we look at the integral and here, let's go ahead into a use of we'LL have to be a little bit careful here because right now and also this one becomes hopes this one over here becomes natural Aga One which is zero and then we have If I write you do you That gives me ln X over x t x But I need extra the p So we have X So we need another excellent p minus one factor here. And here's how we'Ll obtain that so exponentially a both sides of this and then raise both sides of the P minus one And there we go. So we'LL put that instead of writing X to the P minus one in the denominator we'Ll write it in this form over here Now we've ensured that we haven't changed the integral and let me just go ahead and write Those says just put a negative on that p minus one. Now this is the integral that will deal with so maybe write that Okay, so this is an improper integral so we should probably rewrite this is a limit before we integrate. Okay, so well, the older women at the Barre and now we have to integrate this on the inside and here argues integration by parts so using parts here will give us the following and then we have our limits. Zero kay. So then the next step is just the garden plug in those lower and upper limits and zero in K and for you. So that's when you plug in K. Then go ahead and plug in zero the first term. Just become zero so clearly we can see that we should not be taking so far. We are considering Pena equals one. Otherwise thes denominators wouldn't make any sense, so P equals One will deal with that in a moment. But let's not worry about this case right now. So right now we're assuming we're in this case so that all these fractions are defined. Now let's consider two cases, so P is not equal to one. So we have either ppe us than one or bigger than one. So let's assume, for the time being that it's less than one and see if the inner world emerges. So P is less than one. That implies one minus p is positive. And then so now when you take the Limited's cake goes to infinity. This first term right here we'LL go to Well, Kay's going to infinity and then you have a positive number times infinity. So that's also eat to the infinity. And that was just infinity. So that'LL be diversion on the other hand, a piece larger than one, then one minus piece Negative. And this is a good thing because now this becomes infinity. But now we have e to the minus infinity because of the negative, the one minus t being a negative And this limit if you rewrite this So now right is a p minus one p minus one is positive, just won't supply this by a negative and use low Patel's rule Lopez house rule here because it's infinity over affinity and you will get zero here and similarly, this is either the minus infinity. So that will go to zero. And we have a bunch of real numbers that we add. And this will converge no infinities at the very end, so it will converge. So we're mostly done here Now, in the next page will just have to deal with the case where P equals one. So he is one and this case just go back to Original Siri's. Now, this no is bigger than this some here. So perhaps I could be more formal here, So let me go back. So ended The peaches become Zen now was plugging the first few terms Ellen of one zero and so on. And then let me rewrite this as and you'LL see why I'm point pulling out the three in the second. It's because of the following facts. If N is bigger than or equal to three, then natural log of end will be bigger than equals one. So the Siri's is larger than the harmonic series. So here you this diverges, you could either say harmonic. Siri's orjust used to pee Test with P equals one and therefore, by comparison, test and P diversions if P equals one. So, therefore about Wilby showed on the previous page. The original Siri's Comm urges if and only if he's bigger than one.

Find the positive values of P. For which this serious converges. So I should take the spirit of this and take a look at where this actually decreases. So if we take a look at uh creative by question role, so we've got and Mhm Times one of Grand. Yes. How nine event times P and to the pier by this one. Okay. All over. And P B squared. Ok, so simplifying this, we got into the P this one minus pelon of so peter and P times and few months we're here. So Oliver and peeing and to the Peace square. Okay, I'm not gonna worry about the denominator but I'm going to simplify the top so I have end to the P -1 -1 times one minus P I'm event. Okay. Yeah. And here. Right, so then now for here going to look for a where this is supposed to be less than zero, so it's decreasing. And we can see here that if If P was less than or equal to one that uh huh basically I would not here decreasing. So therefore the only values for which it's decreasing, that's when P is greater than one, and so therefore our answer is P has to be greater than one for this, too coverage.


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